# The Elements of Euclid for the Use of Schools and Colleges/Appendix

APPENDIX.

This Appendix consists of a collection of important propositions which will be found useful, both as affording geometrical exercises, and as exhibiting results which are often required in mathematical investigations. The student will have no difficulty in drawing for himself the requisite figures in the cases where they are not given.

1. *The sum of the squares on the sides of a triangle* *is equal to twice the square on half the base, together with* *twice the square on the straight line which joins the vertex* *to the middle point of the base.*

Let *ABC* be a triangle; and let *D* be the middle point of the base *AB*. Draw *CE* perpendicular to the base

meeting it at *E*; then *B* may be either in *AB* or in *AB* produced.

First, let *B* coincide with *D*; then the proposition follows immediately from I. 47.

Next, let *E* not coincide with *D*; then of the two angles *ADC* and *BDC*, one must be obtuse and one acute.

Suppose the angle *ADC* obtuse. Then, by II. 12, the square on *AC* is equal to the squares on *AD*, *DC*, together with twice the rectangle *AD*, *DE*; and, by II. 13, the square on *BC* together with twice the rectangle *BD*, *DE* is equal to the squares on *BD*, *BC*. Therefore, by Axiom 2, the squares on *AC*, *BC*, together with twice the rectangle *BD*, *BE* are equal to the squares on *AD*, *DB*, and twice the square on *DC*, together with twice the rectangle *AD*, *DE*. But *AD* is equal to *DB*. Therefore the squares on *AC*,*BC* are equal to twice the squares on *AD*, *DC*. 2. *If two chords intersect within a circle, the angle which they include is measured by half the sum of the in- tercepted arcs.*

Let the chords *AB* and *CD* of a circle intersect at *E*; join *AD*.

The angle *AEC* is equal to the angles *ADE*, and *DAE*, by I. 32; that is, to the angles standing on the arcs *AC* and *BD*. Thus the angle *AEC* is equal to an angle at the circumference of the circle standing on the sum of the arcs *AC* and *BD*; and is therefore equal to an angle at the centre of the circle standing on half the sum of these arcs.

Similarly the angle *CEB* is measured by half the sum of the arcs *CB* and *AD*.

3. *If two chords produced intersect without a circle, the angle which they include is measured by half the difference of the intercepted arcs.*

Let the chords *AB* and *CD* of a circle, produced, intersect at *E*; join *AD*.

The angle *ADC* is equal to the angles *EAD* and *AED*, by I. 32. Thus the angle *AEC* is equal to the difference of the angles *ADC* and *BAD*; that is, to an angle at the *circumference* of the circle standing on an arc which is the
difference of *AC* and *BD*; and is therefore equal to an angle at the *centre* of the circle standing on half the difference of these arcs. 4. *To draw a straight line which shall touch two given circles.*

Let *A* be the centre of the greater circle, and *B* the centre of the less circle. With centre *A*, and radius equal to the difference of the radii of the given circles, describe a circle; from *B* draw a straight line touching the circle

so described at *C*. Join *AC* and produce it to meet the circumference at *D*. Draw the radius *BE* parallel to *AD* and on the same side of *AB*; and join *DE*. Then *DE* shall touch both circles.

See I. 33, I. 29, and III. 16 Corollary.

Since two straight lines can be drawn from *B* to touch the described circle, two solutions can be obtained; and the two straight lines which are thus drawn to touch the two given circles can be shewn to meet *AB*, produced through *B*, at the same point. The construction is applicable when each of the given circles is without the other, and also when they intersect.

When each of the given circles is without the other we can obtain two other solutions. For, describe a circle with *A* as a centre and radius equal to the sum of the radii of the given circles; and continue as before, except that *BE* and *AD* will now be on *opposite* sides of *AB*. The two straight lines which are thus drawn to touch the two given circles can be shewn to intersect *AB* at the same point. 5. *To describe a circle which shall pass through three given points not in the same straight line.*

This is solved in Euclid IV. 5.

6. *To describe a circle which shall pass through two given points on the same side of a given straight line, and touch that straight line.*

Let *A* and *B* be the given points; join *AB* and produce it to meet the given straight line at *C*. Make a square equal to the rectangle *CA*, *CB* (II. 14), and on the
given straight line take *CE* equal to a side of this square.

Describe a circle through *A*, *B*, *E* (5); this will be the circle required (III. 37).

Since *E* can be taken on either side of *C*, there are two solutions.

The construction fails if *AB* is parallel to the given straight line. In this case bisect *AB* at *D*, and draw *DC* at right angles to *AB*, meeting the given straight line at *C*.

Then describe a circle through *A*, *B*, *C*.

7. *To describe a circle which shall pass through a given point and touch two given straight lines.*

Let *A* be the given point; produce the given straight lines to meet at *B*, and join *AB*. Through *B* draw a straight line, bisecting that angle included by the given straight lines within which *A* lies; and in this bisecting straight line take any point *C*. From *C* draw a perpendicular on one of the given straight lines, meeting it at 'D*; with centre *C*, and radius *CD*, describe a circle, meeting *AB*, produced if necessary, at *E*. Join *CE*; and through *A* draw a straight line parallel to *CE*, meeting *BC*, produced if *
necessary, at *F*. The circle described from the centre *F*, with radius *FA*, will touch the given straight lines.

For, draw a perpendicular from *F* on the straight line *BD*, meeting it at *G*. Then *CE* is to *FA* as *BC* is to *BF*, and *CD* is to *FG* as *BC* is to *BF* (VI. 4, V. 16). Therefore *CE* is to *FA* as *CD* is to *FG* (V. 11). Therefore *CE* is to *CD* as *FA* is to *FG* (V. 16). But *CE* is equal to *CD*; therefore *FA* is equal to *FG* (V. A).

If *A* is on the straight line *BC* we determine *E* as before; then join *ED*, and draw a straight line through *A* parallel to *ED* meeting *BD* produced if necessary at *G*; from *G* draw a straight line at right angles to *BG*, and the point of intersection of this straight line with *BC* produced if necessary, is the required centre.

As the circle described from the centre *C*, with the radius *CD*, will meet *AB* at two points, there are two solutions.

If *A* is on one of the given straight lines, draw from *A* a straight line at right angles to this given straight line; the point of intersection of this straight line with either of the two straight lines which bisect the angles made by the given straight lines may be taken for the centre of the required circle.

If the two given straight lines are parallel, instead of drawing a straight line *BC* to bisect the angle between them, we must draw it parallel to them, and equidistant from them. 8. *To describe a circle which shall touch three given straight lines, not more than two of which are parallel.*

Proceed as in Euclid IV. 4. If the given straight lines form a triangle, four circles can be described, namely, one as in Euclid, and three others each touching one side of the triangle and the other two sides produced. If two of the given straight lines are parallel, two circles can be described, namely, one on each side of the third given straight line.

9. *To describe a circle which shall touch a given circle, and touch a given straight line at a given point.*

Let *A* be the given point in the given straight line, and *C* be the centre of the given circle. Through *C* draw a straight line perpendicular to the given straight line,

and meeting the circumference of the circle at *B* and *D*, of which *D* is the more remote from the given straight line. Join *AD*, meeting the circumference of the circle at *E*. From *A* draw a straight line at right angles to the given straight line, meeting *CE* produced at *F*. Then *F* shall be the centre of the required circle, and *FA* its radius.

For the angle *AEF* is equal to the angle *CED* (I. 15); and the angle *EAF* is equal to the angle *CDE* (I. 29); therefore the angle *AEF* is equal to the angle *EAF*; therefore *AF* is equal to *EF* {I. 6). In a similar manner another solution may be obtained by joining *AB*. If the given straight line falls without the given circle, the circle obtained by the first solution touches the given circle externally, and the circle obtained by. the second solution touches the given circle internally. If the given straight line cuts the given circle, both the circles obtained touch the given circle externally.

10. *To describe a circle which shall pass through two given points and touch a given circle.*

Let *A* and *B* be the given points. Take any point *C* on the circumference of the given circle, and describe a circle through *A*, *B*, *C*. If this described circle touches the given circle, it is the required circle. But if not, let *D*
be the other point of intersection of the two circles. Let *AB* and *CD* be produced to meet at *E*; from *E* draw a straight line touching the given circle at *F*. Then a circle described through *A*, *B*, *F* shall be the required circle.

See III. 35 and III. 37.

There are two solutions, because two straight lines can be drawn from *E* to touch the given circle.

If the straight line which bisects *AB* at right angles passes through the centre of the given circle, the construction fails, for *AB* and *CD* are parallel. In this case *F* must be determined by drawing a straight line parallel to *AB* so as to touch the given circle. 11. *To describe a circle which shall touch two given straight lines and a given circle.*

Draw two straight lines parallel to the given straight lines, at a distance from them equal to the radius of the given circle, and on the sides of them remote from the centre of the given circle. Describe a circle touching the straight lines thus drawn, and passing through the centre of the given circle (7). A circle having the same centre as the circle thus described, and a radius equal to the excess of its radius over that of the given circle, will be the required circle.

Two solutions will be obtained, because there are two solutions of the problem in 7; the circles thus obtained touch the given circle externally.

We may obtain two circles which touch the given circle internally, by drawing the straight lines parallel to the given straight lines on the sides of them adjacent to the centre of the given circle.

12. *To describe a circle which shall pass through a given point and touch a given straight line and a given circle.*

We wall suppose the given point and the given straight line without the circle; other cases of the problem may be treated in a similar manner.

Let *A* be the given point, and *B* the centre of the given circle. From *B* draw a perpendicular to the given straight line, meeting it at *C*, and meeting the circumference of the given circle at *D* and *E*, so that *D* is between *B* and *C*. Join *EA* and determine a point *F* in *EA* produced if necessary, such that the rectangle *EA*, *EF* may be equal to the rectangle *EC*,*ED*; this can be done by describing a circle through *A*,*C*,*D*, which will meet *EA* at the required point (III. 36, Corollary). Describe a circle to pass through *A* and *F* and touch the given straight line (6); this shall be the required circle. For, let the circle thus described touch the given straight line at *G*; join *EG* meeting the given circle at *H*,
and join *DH*. Then the triangles *EHD* and *EGG* are similar; and therefore the rectangle *EC*, *ED* is equal to the rectangle *EG*, *EH* (III. 31, VI. 4, VI. 16). Thus the rectangle *EA*, *EF* is equal to the rectangle *EH*, *EG*; and therefore *H* is on the circumference of the described circle (III. 36, Corollary). Take *K* the centre of the described circle; join *KG*, *KH*, and *BH*. Then it may be shewn that the angles *KHG* and *EHB* are equal (I. 29, I. 5). Therefore *KHB* is a straight line; and therefore the described circle touches the given circle.

Two solutions will be obtained, because there are two solutions of the problem in 6; the circles thus described touch the given circle externally.

By joining *DA* instead of *EA* we can obtain two solutions in which the circles described touch the given circle internally. 13. *To describe a circle which shall touch a given straight line and two given circles.*

Let *A* be the centre of the larger circle and *B* the centre of the smaller circle. Draw a straight line parallel to the given straight line, at a distance from it equal to the radius of the smaller circle, and on the side of it remote from *A*. Describe a circle with *A* as centre, and radius equal to the difference of the radii of the given circles. Describe a circle which shall pass through *B*, touch externally the circle just described, and also touch the straight line which has been drawn parallel to the given straight line (12). Then a circle having the same centre as the second described circle, and a radius equal to the excess of its radius over the radius of the smaller given circle, will be the required circle.

Two solutions will be obtained, because there are two solutions of the problem in 12; the circles thus described touch the given circles externally.

We may obtain in a similar manner circles which touch the given circles internally, and also circles which touch one of the given circles internally and the other externally.

14. *Let *A* be the centre of a circle, and *B* the centre of a larger circle; let a straight line he drawn touching the former circle at *C* and the latter circle at *D*, and meeting *AB* produced through *A* at *T*. From *T* draw any straight line meeting the smaller circle at *K* and *L*, and the larger circle at *M* and *N*; so that the five letters *T*, *K*, *L*, *M*, *N* are in this order. Then the straight lines *AK*, *KC*, *CL*, *LA* shall he respectively parallel to the straight lines *BM*, *MD*, *DN*, *NB*; and the rectangle *TK*, *TN* shall he equal to the rectangle *TL*, *TM*, and equal to the rectangle *TC*, *TD*.*

Join *AC*, *BD*. Then the triangles *TAC* and *TBD* are equiangular; and therefore *TA* is to *TB* as *AC* is to *BD* (VI. 4, V. 16), that is, as *AK* is to *BM*.

Therefore the triangles *TAK* and *TBM* are similar (YI. 7); therefore the angle *TAK* is equal to the angle *TBM*; and therefore *AK* is parallel to *BM*. Similarly *AL* is parallel to *BN*. And because *AK* is parallel to *BM* and *AG* parallel to *BD* the angle *CAK* is equal to the angle *DBM*; and therefore the angle *CLK* is equal to the angle *DNM* (III. 20); and therefore *CL* is parallel to *DN*. Similarly *CK* is parallel to *DM*.

Now *TM* is to *TD* as *TD* is to *TN* (III. 37, VI. 16); and *TM* is to *TD* as *TK* is to *TC* (VI. 4); therefore *TK* is to *TC* as *TD* is to *TN*; and therefore the rectangle *TK*, *TN* is equal to the rectangle *TC*, *TD*, Similarly the rectangle *TL*, *TM* is equal to the rectangle *TC*, *TD*.

If each of the given circles is without the other we may suppose the straight line which touches both circles to meet *AB* at *T* *between* *A* and *B*, and the above results will all hold, provided we interchange the letters *K* and *L*; so that the five letters are now to be in the following order, *L*, *K*, *T*, *M*, *N*

The point T is called a *centre of similitude* of the two circles. 15. *To describe a circle which shall pass through a given point and touch two given circles.*

Let *A* be the centre of the smaller circle and *B* the centre of the larger circle; and let *E* be the given point.

Draw a straight line touching the former circle at *G* and the latter at *D*, and meeting the straight line *AB*, produced through *A*, at *T*. Join *TE* and divide it at *F* so that the rectangle *TE*, *TF* may be equal to the rectangle *TC*, *TD*. Then describe a circle to pass through *E* and *F* and touch either of the given circles (10); this shall be the required circle. For suppose that the circle is described so as to touch the smaller given circle; let *G* be the point of contact; we have then to shew that the described circle will also touch the larger given circle. Join *TG*, and produce it to meet the larger given circle at *H*. Then the rectangle *TG*,*TH* is equal to the rectangle *TC*, *TD* (14); therefore the rectangle *TG*, *TH* is equal to the rectangle *TE*, *TF* and therefore the described circle passes through *H*.

Let *O* be the centre of this circle, so that *OGA* is a straight line; we have to shew that *OHB* is a straight line.

Let *TG* intersect the smaller circle again at *K*; then *AK* is parallel to *BH*(14); therefore the angle *AKT* is equal to the angle *BHG*; and the angle *AKG* is equal to the angle *AGK*, which is equal to the angle *OGH*, which is equal to the angle *OHG*. Therefore the angles *BHG* and *OHG* together are equal to *AKT* and *AKG* together; that is, to two right angles. Therefore *OHB* is a straight line.

Two solutions will be obtained, because there are two solutions of the problem in *10*. Also, if each of the given circles is without the other, two other solutions can be obtained by taking for *T* the point between *A* and *B* where a straight line touching the two given circles meets *B*. The various solutions correspond to the circumstance that the contact of circles may be external or internal

16. *To describe a circle which shall touch three given circles.*

Let *A* be the centre of that circle which is not greater than either of the other circles; let *B* and *C* be the centres of the other circles. With centre *B*, and radius equal to the excess of the radius of the circle with centre *B* over the radius of the circle with centre *A*, describe a circle. Also with centre *C*, and radius equal to the excess of the radius of the circle with centre *C* over the radius of the circle with centre *A*, describe a circle. Describe a circle to touch externally these two described circles and to pass through *A* (15). Then a circle having the same centre as the last described circle, and having a radius equal to the excess of its radius over the radius of the circle with centre *A* will touch externally the three given circles.

In a similar way we may describe a circle touching internally the three given circles, or touching one of them externally and the two others internally, or touching one of them internally and the two others externally.

17. *In a given indefinite straight line it is required to find a point such that the sum of its distances from two given points on the same side of the straight line shall be the least possible.*

Let *A* and *B* be the two given points. From *A* draw a perpendicular to the given straight line meeting it at *C* and produce *AC* to *D* so that *CD* may be equal to *AC* Join *DB* meeting the given straight line at *E*. Then shall be the required point.

For, let *F* be any other point in the given straight line. Then, because *AG* is equal to *DC*, and *EC* is common to the two triangles *ACE*, *DCE*; and that the right angle *ACE* is equal to the right angle *DCE*; therefore *AE* is equal to *DE*. Similarly, *AF* is equal to *DF*. And the sum of *DF* and *FB* is greater than *BD* (I. 20): therefore the sum of *AF* and *FB* is greater than *BD*; that is, the sum of *AF* and *FB* is greater than the sum of *DE* and *EB*; therefore the sum of *AF* and *FB* is greater than the sum of *AE* and *EB*. 18. *The perimeter of an isosceles triangle is less than that of any other triangle of equal area standing on the same base.*
Let *ABC* be an isosceles triangle; *AQC* any other triangle equal in area and standing on the same base *AC*.

Join *BQ*; then *BQ* is parallel to *AC* (I. 39).

And it will follow from 17 that the sum of *AQ* and *QC* is greater than the sum of *AB* and *BC*.

19. *If a polygon he not equilateral a polygon may he found of the same number of sides, and equal in area, but having a less perimeter.*

For, let *CD*, *DE* be two adjacent unequal sides of the polygon. Join *CE*. Through *D* draw a straight line parallel to *CE*. Bisect *CE* at *L*; from *L* draw a straight line at right angles to *CE* meeting the straight line drawn through *D* at *K*. Then by removing from the given polygon the triangle *CDE* and applying the triangle *CKE* we obtain a polygon having the same number of sides as the given polygon, and equal to it in area, but having a less perimeter (18). 20. * *A* and *B* are two given points on the same side of a given straight line, and *AB* produced meets the given straight line at *C*; of all points in the given straight line on each side of *C*, it is required to determine that at which *AB* subtends the greatest angle*.

Describe a circle to pass through *A* and *B*, and to touch the given straight line on that side of *C* which is to be considered (6). Let *D* be the point of contact: *D* shall be the required point.

For, take any other point *E* in the given straight line, on the same side of *C* as *D* is; draw *EA*, *EB*; then one at least of these straight lines will cut the circumference *ADB*.

Suppose that *BE* cuts the circumference at *F*; join *AF*. Then the angle *AFB* is equal to the angle *ADB* (III. 21); and the angle *AFB* is greater than the angle *AEB* (I. 16); therefore the angle *ADB* is greater than the angle *AEB*.

21. * *A* and *B* are two given points within a circle; and *AB* is drawn and produced both ways so as to divide the whole circumference into two arcs; it is required to determine the point in each of these arcs at which *AB* subtends the greatest angle.* Describe a circle to pass through *A* and *B* and to touch the circumference considered (10): the point of contact will be the required point. The demonstration is similar to that in the preceding proposition.

22.* *A* and *B* are two given points without a given circle; it is required to determine the points on the cir- cumference of the given circle at which *AB* subtends the greatest and least angles.*

Suppose that neither *AB* nor *AB* produced cuts the given circle.

Describe two circles to pass through *A* and *B*, and to touch the given circle (10): the point of contact of the circle which touches the given circle externally will be the point where the angle is greatest, and the point of contact of the circle which touches the given circle internally will be the point where the angle is least. The demonstration is similar to that in 20.

If *AB* cuts the given circle, both the circles obtained by 10 touch the given circle internally; in this case the angle subtended by *AB* at a point of contact is less than the angle subtended at any other point of the circumference of the given circle which is on the same side of *AB*. Here the angle is greatest at the points where *AB* cuts the circle, and is there equal to two right angles.

If *AB* produced cuts the given circle, both the circles obtained by 10 touch the given circle externally; in this case the angle subtended by *AB* at a point of contact is greater than the angle subtended at any other point of the circumference of the given circle which is on the same side of *AB*. Here the angle is least at the points where *AB* produced cuts the circle, and is there zero. 23. *If there be four magnitudes such that the first is to the second as the third is to the fourth; then shall the first together with the second be to the excess of the first above the second as the third together with the fourth is to the excess of the third above the fourth.*

For, the first together with the second is to the second as the third together with the fourth is to the fourth (V. 18). Therefore, alternately, the first together with the second is to the third together with the fourth as the second is to the fourth (V. 16).

Similarly, by V. 17 and V. 16, the excess of the first above the second is to the excess of the third above the fourth as the second is to the fourth.

Therefore, by V. 11, the first together with the second is to the excess of the first above the second as the third together with the fourth is to the excess of the third above the fourth.

24. *The straight lines drawn at right angles to the sides of a triangle from the points of bisection of the sides meet at the same point.*

Let *ABC* be a triangle; bisect *BC* at *D*, and bisect *CA* at *E*; from *D* draw a straight line at right angles to *BC*, and from *E* draw a straight line at right angles to *CA*;

let these straight lines meet at *G*: we have then to shew that the straight line which bisects *AB* at right angles also passes through *G*. From the triangles *BDG* and *CDG* we can shew that *BG* is equal to *CG*; and from the triangles *CEG* and *AEG* we can shew that *CG* is equal to *AG*; therefore *BG* is equal to *AG*. Then if we draw a straight line from *G* to the middle point of *AB* we can show that this straight line is at right angles to *AB*: that is, the line which bisects *AB* at right angles passes through *G*.

25. *The straight lines drawn from the angles of a triangle to the points of bisection of the opposite sides meet at the same point.*

Let *ABC* be a triangle; bisect *BC* at *D*, bisect *CA* at *E*, and bisect *AB* at *F*; join *BE* and *CF* meeting at *G*;
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join *AG* and *GD*: then *AG* and *GD* shall lie in a straight line.

The triangle *BEA* is equal to the triangle *BEC*, and the triangle *GEA* is equal to the triangle *GEC* (I. 38); therefore, by the third Axiom, the triangle *BGA* is equal to the triangle *BGC*.

Similarly, the triangle *CGA* is equal to the triangle *CGB*.

Therefore the triangle *BGA* is equal to the triangle *CGA*. And the triangle *BGB* is equal to the triangle *CGB* (1. 38); therefore the triangles *BGA* and *BGD* together are equal to the triangles *CGA* and *CGD* together. Therefore the. triangles *BGA* and *BGD* together are equal to half the. triangle *ABC*. Therefore *G* must fall on the straight line *AB*; that is, *AG* and *GD* lie in a straight line. 26. *The straight lines which bisect the angles of a triangle meet at the same point.*

Let *ABC* be a triangle; bisect the angles at *B* and *C*
by straight lines meeting at *G*; join *AG*: then *AG* shall bisect the angle at *A*.

From *G* draw *GD* perpendicular to *BC*, *GE* perpendicular to *CA*, and *GF* perpendicular to *AB*.

From the triangles *BGF* and *BGD* we can shew that *GF* is equal to *GD*, and from the triangles *CGE* and *CGD* we can shew that *GE* is equal to *GD*; therefore *GF* is equal to *GE*. Then from the triangles *AFG* and *AEG* we can shew that the angle *FAG* is equal to the angle *EAG*.

The theorem may also be demonstrated thus. Produce *AG* to meet *BC* at *H*. Then *AB* is to *BH* as *AG* is to *GH*, and *AC* is to *CH* as *AG* is to *GH* (VI. 3); therefore *AB* is to *BH* as *AC* is to *CH'(V.11); therefore *AB* is to *AC* as *BH* is to *CH* (V. 16); therefore the straight line *AH* bisects the angle at *A* (VI. 3).*

27. *Let two sides of a triangle he produced through the base; then the straight lines which bisect the two exterior angles thus formed, and the straight line which bisects the vertical angle of the triangle, meet at the same point.*

This may be shewn like 26: if we adopt the second method we shall have to use VI. *A*. 28. *The perpendiculars drawn from the angles of a triangle on the opposite sides meet at the same point.*

Let *ABC* be a triangle; and first suppose that it is not obtuse angled. From *B* draw *BE* perpendicular to *CA*;

from *C* draw *CF* perpendicular to *AB*; let these perpendiculars meet at *G*; join *AG*, and produce it to meet *BC* at *D*: then *AD* shall be perpendicular to *BC*.

For a circle will go round *AEGF* {*Note on* III. 22); therefore the angle FAG is equal to the angle *FEG* (III. 21). And a circle will go round *BCEF* (IV. 31, Note on III. 21); therefore the angle *FEB* is equal to the angle *FCB*. Therefore the angle *BAD* is equal to the angle *BCF*. And the angle at *B* is common to the two triangles *BAD* and *BCF*. Therefore the third angle *BDA* is equal to the third angle *BFC* (Note on I. 32). But the angle *BFC* is a right angle, by construction; therefore the angle *BDA* is a right angle.

In the same way the theorem may be demonstrated when the triangle is obtuse angled. Or this case may be deduced from what has been already shewn. For suppose the angle at *A* obtuse, and let the perpendicular from *B* on the opposite side meet that side *produced* at *E*, and let the perpendicular from *C* on the opposite side meet that side *produced* at *F*; and let *BE* and *CF* be produced to meet at *G*. Then in the triangle *BGG* the perpendiculars *BF* and *CE* meet at *A*; therefore by the former case the straight line *GA* produced will be perpendicular to *BC*. 29. *If from any point in the circumference of the circle described round a triangle perpendiculars be drawn to the sides of the triangle, the three points of intersection are in the same straight line.*

Let *ABC* be a triangle, *P* any point on the circumference of the circumscribing circle; from *P* draw *PD*,

*PE*,*PF* perpendiculars to the sides *BC*, *CA*,*AB* respectively: *D*, *B*, *F* shall be in the same straight line.

[We will suppose that *P* is on the arc cut off by *AB*, on the opposite side from *C*, and that *B* is on *CA* produced through *A*; the demonstration will only have to be slightly modified for any other figure.]

A circle will go round *PEAF* (*Note on* III. 22); therefore the angle *PFE* is equal to the angle *PAE* (III. 21). But the angles *PAE* and *PAC* are together equal to two right angles (I. 13); and the angles *PAC* and *PBC* are together equal to two right angles (III. 22). Therefore the angle *PAB* is equal to the angle *PBC*; therefore the angle *PFE* is equal to the angle *PBC*.

Again, a circle will go round *PFDB* {Note on III. 21); therefore the angles *PFD* and *PBD* are together equal to two right angles (III. 22). But the angle *PBD* has been shewn equal to the angle *PFE*. Therefore the angles *PFD* and *PFE* are together equal to two right angles. Therefore *EF* and *FD* are in the same straight line. 30. * *ABC* is a triangle, and *O* is the point of inter-section of the perpendiculars from *A*, *B*, *C* on the opposite sides of the triangle: the circle which passes through the middle points of *A, *OB*, *OC* will pass through the feet of the perpendiculars and through the middle points of the sides of the triangle.

Let *D*, *E*, *F* be the middle points of *OA*, *OB*, *OC* respectively; let *G* be the foot of the perpendicular from *A* on *BC*, and *H ' 'the middle point of *BC*.*

Then *OBC* is a right-angled triangle and *E* is the middle point of the hypotenuse *OB*; therefore *EG* is equal to *EO*; therefore the angle *EGO* is equal to the angle *EOG*. Similarly, the angle *FGO* is equal to the angle *FOG*. Therefore the angle *FGE* is equal to the angle *FOE*. But the angles *FOE* and *BAG* are together equal to two right angles; therefore the angles *FGE* and *BAC* are together equal to two right angles. And the angle *BAC* is equal to the angle *EDF*, because *ED*, *DF* are parallel to *BA*, *AG* (VI. 2). Therefore the angles *FGE* and *EDF* are together equal to two right angles. Hence *G* is on the circumference of the circle which passes through *D*, *E*, *F*
{*Note on* III. 22).

Again, *FH* is parallel to *OB*, and *EH* parallel to *0C* therefore the angle *EHF* is equal to the angle *EGF*. Therefore *H* is also on the circumference of the circle. Similarly, the two points in each of the other sides of the triangle *ABC* may be shewn to be on the circumference of the circle.

The circle which is thus shewn to pass through these nine points may be called the *Nine points circle:* it has some curious properties, of which we will now give two.

*The radius of the Nine points circle is half of the radius of the circle described round the original triangle.*

For the triangle *DEF* has its sides respectively halves of the sides of the triangle *ABC*, so that the triangles are similar. Hence the radius of the circle described round *DEF* is half of the radius of the circle described round *ABC*.

*If *S* be the centre of the circle described round the triangle *ABC*, the centre of the Nine points circle is the middle point of*SO.

For *HS* is at right angles to *BC*, and therefore parallel to *GO*. Hence the straight line which bisects *HG* at right angles must bisect *SO*. And *H* and *G* are on the circumference of the Nine points circle, so that the straight line which bisects *HG* at right angles must pass through the centre of the Nine points circle. Similarly, from the other sides of the triangle *ABC* two other straight lines can be obtained, which pass through the centre of the Nine points circle and also bisect *SO*. Hence the centre of the Nine points circle must coincide with the middle point of *SO*.

We may state that the Nine points circle of any triangle touches the inscribed circle and the escribed circles of the triangle: a demonstration of this theorem will be found in the *Plane Trigonometry*, Chapter xxiv. For the history of the theorem see the *Nouvelles Annales de Mathematiques* for 1863, page 562.

31. If two straight lines bisecting two angles of a triangle and terminated at the opposite sides be equal, the bisected angles shall be equal.

Let *ABC* be a triangle; let the straight line *BD* bisect the angle at *B*, and be terminated at the side *AC*; and let the straight line *CE* bisect the angle at *C*, and be terminated at the side *AB*; and let the straight line *BD* be equal to the straight line *CE*: then the angle at *B* shall be equal to the angle at *C*

For, let *BB* and *GE* meet at; then if the angle *OBC* be not equal to the angle *OCB* one of them must be greater than the other; let the angle *OBC* be the greater. Then, because *CB* and *BD* are equal to *BC* and *CE*, each to each; but the angle *CBD* is greater than the angle *BCE*; therefore *CD* is greater than *BE* (I. 24).

On the other side of the base *BC* make the triangle *BCF* equal to the triangle *CBE*, so that *BF* may be equal to *CE*, and *CF* equal to *BE* (I. 22); and join *DF*.

Then because *BF* is equal to *BD* the angle *BED* is equal to the angle *BDF*. And. the angle *OCD* is, by hypothesis, less than the angle *OBE*; and the angle *COD* is equal to the angle *BOE*; therefore the angle *ODC* is greater than the angle *OEB* (I, 32), and therefore the angle *ODC* is greater than the angle *BFC*.

Hence, by taking away the equal angles *BDF* and *BFD*, the angle *FDC* is greater than the angle *DFC*; and therefore *CF* is greater than *CD* (I. 1 9); therefore *BE* is greater than *CD*.

But it was shewn that *CD* is greater than *BE* which is absurd.

Therefore the angles *OBC* and *OCB* are not unequal, that is, they are equal; and therefore the angle *ABC* is equal to the angle *ACB*.

[For the history of this theorem see *Lady's and Gentleman's Diary* for 1869, page 88.] 32. *If a quadrilateral figure does not admit of having a circle described round it, the sum of the rectangles contained by the opposite sides is greater than the rectangle contained by the diagonals.*

Let *ABCD* be a quadrilateral figure which does not admit of having a circle described round it; then the rectangle *AB*, *DC*, together with the rectangle *BC*, *AD*, shall be greater than the rectangle *AC*, *BD*.

For, make the angle *ABE* equal to the angle *DBC*, and the angle *BAE* equal to the angle *BDC*; then the triangle *ABE* is similar to the triangle *BDC* (VI. 4); therefore *AB* is to *AE* as *DB* is to *DC*; and therefore the rectangle *AB*, *DC* is equal to the rectangle *AE*, *DB*.

Join *EC*. Then, since the angle *ABE* is equal to the angle *DBC*, the angle *CBE* is equal to the angle *DBA*. And because the triangles *ABE* and *DBC* are similar, *AB* is to *DB* as *BE* is to *BC*; therefore the triangles *ABD* and *EBC* are similar (VI. 6); therefore *CB* is to *CE* as *DB* is to *DA*; and therefore the rectangle *CB*, *DA* is equal to the rectangle *CE*, *DB*.

Therefore the rectangle *AB*, *DC*, together with the rectangle *BC*, *AD* is equal to the rectangle *AE*, *BD* together with the rectangle *CE*, *BD*; that is, equal to the rectangle contained by *BD* and the sum of *AE* and *EC*. But the sum of *AE* and *EC* is greater than *AC* (I. 20); therefore the rectangle *AB*, *DC*, together with the rectangle *BC*, *AD* is greater than the rectangle *AC*, *BD*. 33. *If the rectangle contained by the diagonals of a quadrilateral he equal to the sum of the rectangles con-tained by the opposite sides, a circle can be descrihed round the quadrilateral.*

This is the converse of VI. *D*; it can be demonstrated indirectly with the aid of 32.

34. *It is required to find a point in a given straight line, such that the rectangle contained hy its distances from two given points in the straight line may he equal to the rectangle contained by its distances from two other given points in the straight line.*

Let *A*, *B*, *C*, *D* be four given points in the same straight line: it is required to find a point in the straight
line, such that the rectangle contained by its distances from *A* and *B* may be equal to the rectangle contained by its distances from *C* and *D*.

On *AD* describe any triangle *AED*; and on *CB* describe a similar triangle *CFB*, so that *CF' is parallel to *AE*, and *BF* to *DE*; join *EF*, and let it meet the given straight line at *O*, Then shall be the required point.*

For, *OE* is to *OA* as *OE* is to *OC* (VI. 4); therefore *OE* is to *OF* as *OA* is to *OC* (V. 16). Similarly *OE* is to *OF* as *OD* is to *OB*. Therefore *OA* is to *OC* as *OD* is to *OB* (V. 11). Therefore the rectangle *OA*, *OB* is equal to the rectangle *OC*, *OD*. The figure will vary slightly according to the situation of the four given points, but corresponding to an assigned situation there will be only *one* point such as is required. For suppose there could be such a point *P*, besides the point *O* which is determined by the construction given above; and that the points are in the order *A*,*C*,*D*, *B*, *O*, *P*. Join *PE*, and let it meet *CF*, produced at *G*; join *BG*. Then the rectangle *PA*, *PB* is, by hypothesis, equal to the rectangle *PC*, *PD*; and therefore *PA* is to *PC* as *PD* is to *PB*. But *PA* is to *PC* as *PE* is to *PG* ( VI. 2); therefore *PD* is to *PB* as *PE* is to *PG* (V. 11); therefore BG is parallel to DE.

But, by the construction, *BF* is parallel to *ED*; therefore *BG* and *BF* are themselves parallel (I, 30); which is absurd. Therefore *P* is not such a point as is required.

35. The substantives *analysis* and *synthesis*, and the corresponding adjectives *analytical* and *synthetical*, are of frequent occurrence in mathematics. In general *analysis* means decomposition, or the separating a whole into its parts, and *synthesis* means composition, or making a whole out of its parts. In Geometry however these words are used in a more special sense. In synthesis we begin with results already established, and end with some new result; thus, by the aid of theorems already demonstrated, and problems already solved, we demonstrate some new theorem, or solve some new problem. In analysis we begin with assuming the truth of some theorem or the solution of some problem, and we deduce from the assumption consequences which we can compare with results already established, and thus test the validity of our assumption.

36. The propositions in Euclid's Elements are all exhibited synthetically; the student is only employed in examining the soundness of the reasoning by which each successive addition is made to the collection of geometrical truths already obtained; and there is no hint given as to the manner in which the propositions were originally discovered. Some of the constructions and demonstrations appear rather artificial, and we are thus naturally induced to enquire whether any 'rules can be discovered by which we may be guided easily and naturally to the investigation of new propositions. 37. Geometrical analysis has sometimes been described in language which might lead to the expectation that directions could be given which would, enable a student to proceed to the demonstration of any proposed theorem, or the solution of any proposed problem, with confidence of success; but no such directions can be given. We will state the exact extent of these directions. Suppose that a new theorem is proposed for investigation, or a new problem for trial. Assume the truth of the theorem or the solution of the problem, and deduce consequences from this assumption combined with results which have been already established. If a consequence can be deduced which contradicts some result already established, this amounts to a demonstration that our assumption is inadmissible; that is, the theorem is not true, or the problem cannot be solved. If a consequence can be deduced which coincides with some result already established, we cannot say that the assumption is inadmissible; and it *may happen* that by starting from the consequence which we deduced, and retracing our steps, we can succeed in giving a synthetical demonstration of the theorem, or solution of the problem. These directions however are very vague, because no certain rule can be prescribed by which we are to combine our assumption with results already established; and moreover no test exists by which we can ascertain whether a valid consequence which we have drawn from an assumption will enable us to establish the assumption itself. That a proposition may be false and yet furnish consequences which are true, can be seen from a simple example. Suppose a theorem were proposed for investigation in the following words; *one angle of a triangle is to another as the side opposite to the first angle is to the side opposite to the other.* If this be assumed to be true we can immediately deduce Euclid's result in I. 19; but from Euclid's result in I. 19 we cannot retrace our steps and establish the proposed theorem, and in fact the proposed theorem is false.

Thus the only definite statement in the directions respecting Geometrical analysis is, that if a consequence can be deduced from an assumed proposition which contradicts a result already established, that assumed proposition must be false. 38. We may mention, in particular, that a consequence would contradict results already established, if we could shew that it would lead to the solution of a problem already given up as impossible. There are three famous problems which are now admitted to be beyond the power of Geometry; namely, to find a straight line equal in length to the circumference of a given circle, to trisect any given angle, and to find two mean proportionals between two given straight lines. The grounds on which the geometrical solution of these problems is admitted to be impossible cannot be explained without a knowledge of the higher parts of mathematics; the student of the Elements may however be content with the fact that innumerable attempts have been made to obtain solutions, and that these attempts have been made in vain.

The first of these problems is usually referred to as the *Quadrature of the Circle*. For the history of it the student should consult the article in the English Cyclopædia under that head, and also a series of papers in the *Athenæum* for 1863 and subsequent years, entitled a *Budget of Paradoxes*, by Professor De Morgan.

For *approximate* solutions of the problem we may refer to Davies's edition of Hutton's Course of Mathematics* Vol. i. page 400, the *Lady's and Gentleman's Diary* for 1855, page 86, and the *Philosophical Magazine* for April, 1862.*

The third of the three problems is often referred to as the *Duplication of the Cube*. See the note on VI. 13 in Lardner's *Euclid*, and a dissertation by C. H. Biering entitled Historia Problematis Cubi Duplicandi... Hauniæ, 1844.

We will now give some examples of Geometrical analysis.

39. *From two given points it is required to draw to the same point in a given straight line, two straight lines equally inclined to the given straight line.*

Let *A* and *B* be the given points, and *CD* the given straight line.

Suppose *AE* and *EB* to be the two straight lines equally inclined to *CD*. Draw *BF* perpendicular to *CD*, and produce *AE* and *BF* to meet at *G*. Then the angle *BED* is equal to the angle *AEC* by hypothesis; and the angle *AEC* is equal to the angle *DEG* (I. 15). Hence the
triangles *BEF* and *GEF* are equal in all respects (I. 26); therefore *FG* is equal to *FB*.

This result shews how we may synthetically solve the problem. Draw *BF* perpendicular to *CD*, and produce it to *G*, so that *FG* may be equal to *FB*; then join *AG*, and *AG* will intersect *CD* at the required point.

40. *To divide a given straight line into two parts such that the difference of the squares on the parts may be equal to a given square.*
Let *AB* he the given straight line, and suppose *C* the required point.

Then the difference of the squares on *C* and *BG* is to be equal to a given square. But the difference of the squares on *AC* and *BC* is equal to the rectangle contained by their sum and difference; therefore this rectangle must be equal to the given square. Hence we have the following synthetical solution. On *AB* describe a rectangle equal to the given square (I. 45); then the difference of *AC* and *CB* will be equal to the side of the rectangle adjacent to *AB*, and is therefore known. And the sum of *AC* and *CB* is known. Thus *AC* and *CB* are known.

It is obvious that the given square must not exceed the square on *AB*, in order that the problem may be possible.

There are two positions of *C*, if it is not specified which of the two segments *AC* and *CB* is to be greater than the other; but only one position, if it is specified. In like manner we may solve the problem, *to produce a given straight line so that the square on the whole straight line made up of the given straight line and the part produced, may exceed the square on the part produced by a given square, which is not less than the square on the given straight line*.

The two problems may be combined in one enunciation thus, *to divide a given straight line internally or externally so that the difference of the squares ow the segments may be equal to a given square.*

41. *To find a point in the circumference of a given segment of a circle, so that the straight lines which join the point to the extremities of the straight line on which the segment stands may be together equal to a given straight line.*

Let *ACB* be the circumference of the given segment, and suppose *C* the required point, so that the sum of *AC* and *CB* is equal to a given straight line.

Produce *AC* to *D* so that *CD* may be equal to *CB*; and join *DB*.

Then *AD* is equal to the given straight line. And the angle *ACB* is equal to the sum of the angles *CDB* and *CBD* (I. 32), that is, to twice the angle *CDB* (I. 5). Therefore the angle *ADB* is half of the angle in the given segment. Hence we have the following synthetical solution. Describe on *AB* a segment of a circle containing an angle equal to half the angle in the given segment. With *A* as centre, and a radius equal to the given straight line, describe a circle. Join *A* with a point of intersection of this circle and the segment which has been described; this joining straight line will cut the circumference of the given segment at a point which solves the problem.

The given straight line must exceed *AB* and it must not exceed a certain straight line which we will now determine. Suppose the circumference of the given segment bisected at *E*: join *AE*, and produce it to meet the circumference of the described segment at *F*. Then *AE* is equal to *EB* (III. 28), and *EB* is equal to *EF* for the same reason that *CB* is equal to *CD*. Thus *EA*, *EB*, *EF* are all equal; and therefore *E* is the centre of the circle of which *ADB* is a segment (III. 9). Hence *AF* is the longest straight line which can be drawn from *A* to the circumference of the described segment; so that the given straight line must not exceed twice *AE*,

42.*To describe an isosceles triangle having each of the angles at the base double of the third angle.*

This problem is solved in IV. 10; we may suppose the solution to have been discovered by such an analysis as the following.

Suppose the triangle *ABD* such a triangle as is required, so that each of the angles at *B* and *D* is double of the angle at *A*.

Bisect the angle at *D* by the straight line *DC*. Then the angle *ADC* is equal to the angle at *A*; therefore *CA* is equal to *CD*. The angle *CBD* is equal to the angle *ADB*, by hypothesis; the angle *CDB* is equal to the angle at *A*; therefore the third angle *BCD* is equal to the third angle *ABD* (I. 32). Therefore *BD* is equal to *CD* (I. 6); and therefore *BD* is equal to *AC*.

Since the angle *BDC* is equal to the angle at *A*, the straight line *BD* will touch at *D* the circle described round the triangle *ACD* {*Note on* III. 32). Therefore the rectangle *AB*, *BC* is equal to the square on *BD* (III. 36). Therefore the rectangle *AB*,*BC* is equal to the square on a *AC*

Therefore *AB* is divided at *C* in the manner required in II. 11.

Hence the synthetical solution of the problem is evident. 43. *To inscribe a square in a given triangle.*

Let *ABC* be the given triangle, and suppose *DEFG* the required square. Draw *AH* perpendicular to *EC*, and *AK* parallel to *EG*; and let *EF* produced meet *AK* at *K* Then *BG* is to *GF* as *BA* is to *AK*, and *EG* is to *GD* as *BA* is to AH{VI. 4). But *GF* is equal to *GD*, by hypothesis. Therefore *BA* is to *AK* as *BA* is to *AH* (V. 7, V. 11). Therefore *AH* is equal to *AK*{V. 7).

Hence we have the following synthetical solution. Draw *AK* parallel to *EG*, and equal to *AH*; and join *BK*. Then *BK* meets *AC* at one of the corners of the required square, and the solution can be completed.

44. *Through a given point between two given straight lines, it is required to draw a straight line, such that the rectangle contained by the parts between the given point and the given straight lines may be equal to a given rectangle.*

Let *P* be the given point, and *AE* and *AC* the given straight lines; suppose *MPN* the required straight line, so that the rectangle *MP*, *PN* is equal to a given rectangle.

Produce *AP* to *Q*, so that the rectangle *AP*, *PQ* may be equal to the given rectangle. Then the rectangle *MP*, *PN* is equal to the

rectangle *AP*, *PQ*. Therefore a circle will go round *AMQN* {*Note on* III. .35). Therefore the angle *PNQ* is equal to the angle* PAM* {III. 21).

Hence we have the following synthetical solution. Produce *AP* to *Q*, so that the rectangle *AP*, *PQ* may be equal to the given rectangle; describe on *PQ* a segment of a circle containing an angle equal to the angle *PAM*; join *P* with a point of intersection of this circle and *AC*; the straight line thus drawn solves the problem. 45. *In a given circle it is required to inscribe a tri- angle so that two sides may pass through two given points, and the third side he parallel to a given straight line.*

Let *A* and *B* be the given points, and *CD* the given straight line. Suppose *PMN* to be the required triangle inscribed in the given circle.

Draw *NE* parallel to *AB* join *EM*, and produce it if necessary to meet *AB* at *F*.

If the point *F* were known the problem might be considered solved. For *ENM* is a known angle, and therefore the chord *EM* is known in magnitude. And then, since *F* is a known point, and *EM* is a known magnitude, the position of *M* becomes known.

We have then only to shew how *F* is to be determined. The angle *MEN* is equal to the angle *MFA* (I. 29). The angle *MEN* is equal to the angle *MPN* (III. 21). Hence *MAF* and *BAP* are similar triangles (VI. 4). Therefore *MA* is to *AF* as *BA* is to *AP*. Therefore the rectangle *MA*, *AP* is equal to the rectangle *BA*, *AF*(VI. 16). But since *J* is a given point the rectangle *MA*, *AP* is known; and *AB* is known: thus *AF* is determined. 46. *In a given circle it is required to inscribe a tri-* *angle so that the sides may pass through three given* *points.*

Let *A*, *B*, *C* be the three given points. Suppose *PMN* to be the required triangle inscribed in the given circle

Draw *NE* parallel to *AB*, and determine the point *F* as in the preceding problem. We shall then have to describe in the given circle a triangle *EMN* so that two of its sides may pass through given points, *F* and *C*, and the third side be parallel to a given straight line *AB*. This can be done by the preceding problem.

This example and the preceding are taken from the work of Catalan already cited. The present problem is sometimes called *Castillon's* and sometimes *Cramer's*; the history of the general researches to which it has given rise will be found in a series of papers in the *Mathematician* Vol. Ill, by the late T. S. Davies.

47. *A locus* consists of all the points which satisfy certain conditions and of those points alone. Thus, for example, the locus of the points which are at a given distance from a given point is the surface of the sphere described from the given point as centre, with the given distance as radius; for all the points on this surface, and no other points, are at the given distance from the given point. If we restrict ourselves to all the points in a fixed plane which are at a given distance from a given point, the locus is the circumference of the circle described from the given point as centre, with the given distance as radius. In future we shall restrict ourselves to loci which are situated in a fixed plane, and which are properly called *plane loci*.

Several of the propositions in Euclid furnish good examples of loci. Thus the locus of the vertices of all triangles which are on the same base and on the same side of it, and which have the same area, is a straight line parallel to the base; this is shewn in I. 37 and I. 39.

Again, the locus of the vertices of all triangles which are on the same base and on the same side of it, and which have the same vertical angle, is a segment of a circle described on the base; for it is shewn in III. 21, that all the points thus determined satisfy the assigned conditions, and it is easily shewn that no other points do.

We will now give some examples. In each example we ought to shew not only that all the points which we indicate as the locus do fulfil the assigned conditions, but that no other points do. This second part however we leave to the student in all the examples except the last two; in these, which are more difficult, we have given the complete investigation.

48. *Required the locus of points which are equidistant from two given points.*

Let *A* and *B* be the two given points; join *AB*; and draw a straight line through the middle point of *AB* at right angles to *AB*; then it may be easily shewn that this straight line is the required locus.

49. *Required the locus of the vertices of all triangles on a given base *AB*, such that the square on the side terminated at *A* may exceed the square on the side terminated at *B*, by a given square.*

Suppose *C* to denote a point on the required locus; from *C* draw a perpendicular on the given base, meeting it, produced if necessary, at *D*. Then the square on *AC* is equal to the squares on *AD* and *CD*, and the square on *BC* is equal to the squares on *BD* and *CD* (I. 47); therefore the. square on *AC* exceeds the square on *BC* by as much as the square on *AD* exceeds the square on *BD*. Hence *D* is a fixed point either in *AB* or in *AB* produced through *B* (40). And the required locus is the straight line drawn through *D*, at right angles to *AB*.

50. *Required the locus of a point such that the straight* *lines drawn from it to touch two given circles may be equal*

Let *A* be the centre of the greater circle, *B* the centre of a smaller circle; and let *P* denote any point on the required locus. Since the straight lines drawn from *P* to touch the given circles are equal, the squares on these straight lines are equal. But the squares on *PA* and *PB* exceed these equal squares by the squares on the radii of the respective circles. Hence the square on *PA* exceeds the square on *PB*, by a known square, namely a square equal to the excess of the square on the radius of the circle of which *A* is the centre over the square on the radius of the circle of which *B* is the centre. Hence, the required locus is a certain straight line which is at right angles to *AB* (49).

This straight line is called the radical axis of the two circles.

If the given circles intersect, it follows from III. 36, that the straight line which is the locus coincides with the produced parts of the common chord of the two circles.

51. *Required the locus of the middle points of all the chords of a circle which pass through a fixed point.*

Let *A* be the centre of the given circle; *B* the fixed
point; let any chord of the circle be drawn so that, produced if necessary, it may pass through *B*. Let *P* be the middle point of this chord, so that *P* is a point on the required locus.

The straight line *AP* is at right angles to the chord of which *P* is the middle point (III. 3); therefore *P* is on the circumference of a circle of which *AB* is a diameter. Hence if *B* be within the given circle the locus is the circumference of the circle described on *AB* as diameter; if *B* be without the given circle the locus is that part of the circumference of the circle described on *AB* as diameter, which is within the given circle.

52. O *is a fixed point from which any straight line is drawn meeting a fixed straight line at *P*; in *OP* a point *Q* is taken such that *OQ* is to *OP* in a fixed ratio: determine the locus of *Q*.*

We shall shew that the locus of *Q* is a straight line.

For draw a perpendicular from *O* on the fixed straight line, meeting it at *C*; in *0C* take a point *D* such that *OD* is to *OC* in the fixed ratio; draw from *O* any straight line *OP* meeting the fixed straight line at *P*, and in *OP* take a point *Q* such that *OQ* is to *OP* in the fixed ratio; join

*QD*. The triangles *ODQ* and *OCP* are similar (VI. 6); therefore the angle *ODQ* is equal to the angle *OCP*, and is therefore a right angle. Hence *Q* lies in the straight line drawn through *D* at right angles to *OD*. 53. O *is a fixed point from which any straight line* *is drawn meeting the circumference of a fixed circle at *P* *; *in *OP* a point *Q* is taken such that *OQ* is to *OP* in a fixed* *ratio: determine the locus of *Q*.*

We shall show that the locus is the circumfercuce of a circle.

For let *C* be the centre of the fixed circle; in *OC* take a point *D* such that *OD* is to *OC* in the fixed ratio, and draw any radius *CP* of the fixed circle; draw *DQ* parallel to *CP* meeting *OP*, produced if necessary, at *Q*. Then the triangles *OCP* and *ODQ* are similar (VI. 4), and therefore *OQ* is to *OP* as *OD* is to *0C*, that is, in the fixed ratio. Therefore *Q* is a point on the locus. And *DQ* is to *CP* in the fixed ratio, so that *DQ* is of constant length. Hence the locus is the circumference of a circle of which *D* is the centre.

54. *There are four given points *A*, *B*, *C*, *D* in a straight line; required the locus of a point at which *AB* and *CD* subtend equal angles.*

Find a point *O* in the straight line, such that the rectangle *OA*, *OD* may be equal to the rectangle *OB*, *OC* (34), and take *OK* such that the square on *OK* may be equal to either of these rectangles (II. 14): the circumference of the circle described from as centre, with radius *OK* shall be the required locus.

[We will take the case in which the points are in the following order, *O*, *A*, *B*, *C*, *D*.]

For let *P* be any point on tho circumference of this circle. Describe a circle round *PAD*, and also a circle
round *PBC*; then *OP* touches each of these circles (III. 37); therefore the angle *OPA* is equal to the angle *PDA*, and the angle *OPB* is equal to the angle *PCB* {III. 32). But the angle *OPB* is equal to the angles *OPA* and *APB* together, and the angle *PCB* is equal to the angles *CPD* and *PDA* together (I. 32). Therefore the angles *OPA* and *APB* together are equal to the angles *CPD* and *PDA* together; and the angle *OPA* has been shewn equal to the angle *PDA*; therefore the angle *APB* is equal to the angle *CPD*.

We have thus shewn that any point on the circumference of the circle satisfies the assigned conditions; we shall now shew that any point which satisfies the assigned conditions is on the circumference of the circle.

For take any point *Q* which satisfies the required conditions. Describe a circle round *QAD*, and also a circle round *QBC*. These circles will touch the same straight line at *Q*; for the angles *AQB* and *CQD* are equal, and the converse of III. 32 is true. Let this straight line which touches both circles at *Q* be drawn; and let it meet the straight line containing the four given points at *R*. Then the rectangle *RA*, *RD* is equal to the rectangle *RB*,*RC*; for each is equal to the square on *RQ* (III. 36). Therefore *R* must coincide with *O* (34); and therefore *RQ* must be equal to *OK*. Thus *Q* must be on the circumference of the circle of which is the centre, and *OK* the radius.

55. *Required the locus of the vertices of all the triangles *ABC* which stand on a given base *AB*, and have the side *AC* to the side *BC* in a constant ratio*.

If the sides *AC* and *BC* are to be equal, the locus is
the straight lino which bisects *AB* at right angles. "We will suppose that the ratio is greater than a ratio of equality; so that *AC* is to be the greater side.

Divide *AB* at *D* so that *AD* is to *DB* in the given ratio (VI. 10); and produce *AB* to *E*, so that *AE* is to *EB* in the given ratio. Let *P* be any point in the required locus; join *PD* and *PE*. Then *PD* bisects the angle *APB*, and *PE* bisects the angle between *BP* and *AP* produced. Therefore the angle *DPE* is a right angle. Therefore *P* is on the circumference of a circle described on *DE* as diameter.

We have thus shewn that any point which satisfies the assigned conditions is on the circumference of the circle described on *DE* as diameter; we shall now shew that any point on the circumference of this circle satisfies the assigned conditions.

Let *Q* be any point on the circumference of this circle, *QA* shall be to *QB* in the assigned ratio. For, take *O* the centre of the circle; and join *QO*. Then, by construction, *AE* to *EB* as *AD* is to *DB*, and therefore, alternately, *AE* to *AD* as *EB* is to *DB*; therefore the sum of *AE* and *AD* is to their difference as the sum of *EB* and *DB* is to their difference (23); that is, twice *DO* is to twice *DO* as twice *DO* is to twice *BO*; therefore *AO* is to *DO* as *DO* is to *BO* that is, *AO* is to *OQ* as *QO* is to *OB*. Therefore the triangles *AOQ* and *QOB* are similar triangles (VI. 6); and therefore *AQ* is to *QB* as *QO* is to *BO*. This shews that the ratio of *AQ* to *BQ* is constant; we have still to shew that this ratio is the same as the assigned ratio.

We have already shewn that *AO* is to *DO* as *DO* is to *BO*; therefore, the difference of *AO* and *BO* is to *DO* as the difference of *DO* and *BO* is to *BO* (V. 17); that is, *AD* is to *DO* as *BD* is to *BO*; therefore *AD* is to *BD* as *DO* is to *BO*; that is, *AD* is to *DB* as *QO* is to *BO*. This shews that the ratio of *QO* to *BO* is the same as the assigned ratio.

56. We have hitherto restricted ourselves' to Euclid's Elements, and propositions which can be demonstrated by strict adherence to Euclid's methods. In modern times various other methods have been introduced, and have led to numerous and important results. These methods may be called semi-geometrical, as they are not confined within the limits of the ancient pure geometry; in fact the power of the modern methods is obtained chiefly by combining arithmetic and algebra with geometry. The student who desires to cultivate this part of mathematics may consult Townsend's *Chapters on the Modern Geometry of the Pointy Line, and Circle*.

We will give as specimens some important theorems, taken from what is called the theory of transversals.

Any line, straight or curved, which cuts a system of other lines is called a transversal; in the examples which we shall give, the lines will be straight lines, and the system svill consist of three straight lines forming a triangle.

We will give a brief enunciation of the theorem which we are about to prove, for the sake of assisting the memory in retaining the result; but the enunciation will not be fully comprehended until the demonstration is completed. 57. *If a straight line cut the sides, or the sides produced, of a triangle, the product of three segments in order is equal to the product of the other three segments.*

Let *ABC* be a triangle, and let a straight line be drawn cutting the side *BC* at *D*, the side *CA* at *E*, and the side *AB* produced through *B* at *F*. Then *BD* and *DC* are
called *segments* of the side *BC*, and *CE* and *EA* are called segments of the side *CA*, and also *AF* and *FB* are called *segments* of the side *AB*.

Through *A* draw a straight line parallel to *BC*, meeting *DF* produced at *H*.

Then the triangles *CED* and *EAH* are equiangular to one another; therefore *AH* is to *CD* as *AE* is to *EC* (VI. 4). Therefore the rectangle *AH*, *EC* is equal to the rectangle *CD*, *AE* (VL 16).

Again, the triangles *FAH* and *FBD* are equiangular to one another; therefore *AH* is to *BD* as *FA* is to *FB* (VI. 4). Therefore the rectangle *AH*, *FB* is equal to the rectangle *BD*, *FA* (VI. 16).

Now suppose the straight lines represented by numbers in the manner explained in the notes to the second Book of the Elements. We have then two results which we can express arithmetically: namely, the *prooduct* *AH*.*EC* is equal to the product *CD'.*AE*; and the product *AH*.*FB* is equal to the product *BD*.*FA*.*

Therefore, by the principles of arithmetic, the product *AH*.*EC*.*BD*.*FA* is equal to the product *AH*.*FB*.*CD*.*AE*, and therefore, by the principles of arithmetic, the product *BD*.*CE*.*AF* is equal to the product *DC*.*EA*. *FB*.

This is the result intended by the enunciation given above. Each product is made by three segments, one from every side of the triangle: and the two segments which terminated at any angular point of the triangle are never in the same product. Thus if wo begin one product with the segment *BD*, the other segment of the side *BC*, namely *DC*, occurs in the other product; then the segment *CE* occurs in the first product, so that the two segments *CD* and *CE*, which terminate at *C*, do not occur in the same product; and so on.

The student should for exercise draw another figure for the case in which the transversal meets all the sides produced, and obtain the same result.

58. Conversely, it may be shewn by an indirect proof that if the product *BD*.*CE*.*AF* be equal to the product *DC*.*EA*.*FB*, the three points *D*,*E*,*F* lie in the same straight line.

59. *If three straight lines he drawn through the angular points of a triangle to the opposite sides, and meet at the same pointy the product of three segments in order is equal to the product of the other three segments.*

Let *ABC* be a triangle. From the angular points to the opposite sides let the straight lines *AOD*, *BOE*, *COF* be drawn, which meet at the point *O*: the product *AF*.*BD*.*CE* shall be equal to the product *FB*.*DC*.*EA*.

For the triangle *ABD* is cut by the transversal *FOC*, and therefore by the theorem in 57 the following products are equal, *AF*.'*BC*.*DO*, and *FB*.*CD*.*OA*.

Again, the triangle *ACD* is cut by the transversal *EOB*, and therefore by the theorem in 57 the following products are equal, *AO*.*DB*.*CE* and *OD*.*BC*.*EA*.

Therefore, by the principles of arithmetic, the following products are equal, *AF*.*BC*.*DO*.*AO*.*DB*.*CE* and *FB*.*CD*.*OA*.*OD*.*BC*.*EA*. Therefore the following
products are equal, *AF*.*BD*.*GE* and *FB*.*DC*. *EA*. W have supposed the point to be within the triangle; if *O* be without the triangle two of the points *D*, *E*, *F* will fall on the sides produced.

60. Conversely, it may be shewn by an indirect proof that if the product *AF*. *BD* . *CE* be equal to the product *FB*.*DC*.*EA*, the three straight lines *AD*, *BE*, *CF* meet at the same point.

6l. We may remark that in geometrical problems the following terms sometimes occur, used in the same sense as in arithmetic; namely *arithmetical progression, geometrical progression*, and *harmonical progression*. A proposition respecting harmonical progression, which deserves notice, will now be given.

62. *Let ABC he a triangle; let the angle A be bisected by a straight line which meets BC at D, and let the exterior angle at A be bisected by a straight line which meets BC produced through C, at E: then BD, BC, BE shall he in harmonical progression*.

For *BD* is to *DC* as *BA* is to *AC* (VI. 3); and *BE* is to *AC* as *BA* is to *AC* (VI. A). Therefore *BD* is to *DC* as *BE* is to *EC (V. 11). Therefore *BD* is to *BE* as *DC* is to *EC* (V. 16). Thus of the three straight lines *BD*,*BC*, *BE*, the first is to the third as the excess of the second over the first is to the excess of the third over the second. Therefore *BD*, *BC*, *BE* are in harmonical progression.*

This result is sometimes expressed by saying that *BE* is divided harmonically at *D* and *C*.