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

*BOOK III.*

DEFINITIONS.

1. Equal circles are those of which the diameters are equal, or from the centres of which the straight lines to the circumferences are equal.

This is not a definition, but a theorem, the truth of which is evident; for, if the circles be applied to one another, so that their centres coincide, the circles must likewise coincide, since the straight lines from the centres are equal.

2. A straight line is said to touch a circle, when it meets the circle, and being produced does not cut it.

3. Circles are said to touch one another, which meet but do not cut one another. 4. Straight lines are said to be equally distant from the centre of a circle, when the perpendiculars drawn to them from the centre are equal.

5. And the straight line on which the greater perpendicular falls, is said to be farther from the centre. 6. A segment of a circle is the figure contained by a straight line and the circumference it cuts off.

7. The angle of a segment is that which is contained by the straight line and the circumference.

8. An angle in a segment is the angle contained by two straight lines drawn from any point in the circumference of the segment to the extremities of the straight line which is the base of the segment.

9. And an angle is said to insist or stand on the circumference Intercepted between the straight lines which contain the angle.

10. A sector of a circle is the figure contained by two straight lines drawn from the centre, and the circumference between them.

11. Similar segments of circles are those in which the angles are equal, or which contain equal angles.

[*Note*. In the following propositions, whenever the expression "straight lines from the centre," or "drawn from the centre," occurs, it is to be understood that the lines are drawn to the circumference.

Any portion of the circumference is called an *arc*.]

*PROBLEM*.

*To find the centre of a given circle*.

Let *ABC* be the given circle: it is required to find its centre.
Draw within it any straight line *AB* and bisect *AB* at *D*; [I. 10. from the point *D* draw *DC* at right angles to *AB*; [I. 11. produce *CD* to meet the circumference at *E*, and bisect *CE* at *F*. [I. 10.

The point *F* shall be the centre of the circle *ABC*.

For if *F* be not the centre, if possible, let *G* be the centre; and join *GA*, *GD*, *GB*. Then, because *DA* is equal to *DB*, [*Construction*. and *DG* is common to the two triangles *ADG*, *BDG*; the two sides *AD*, *DG* are equal to the two sides *BD*, *DG*, each to each;

and the base *GA* is equal to the base *GB*, because they are drawn from the centre *G*; [I. *Definition* 15.

therefore the angle *ADG* is equal to the angle *BDG*. [I. 8. But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; [I. *Definition* 10.

therefore the angle *BDG* is a right angle.

But the angle *BDF* is also a right angle. [*Construction*.

Therefore the angle *BDG* is equal to the angle *BDF*, [*Ax*.ll

the less to the greater; which is impossible.

Therefore *G* is not the centre of the circle *ABC*.

In the same manner it may be shewn that no other point out of the line *CE* is the centre;

and since *CE* is bisected at *F*, any other point in *CE*

divides it into unequal parts, and cannot be the centre.

Therefore no point but *F* is the centre;

that is, *F* is the centre of the circle *ABC*:*which was to he found*.

*THEOREM*.

*If any two points he taken in the circumference of a circle, the straight line which joins them shall fall within the circle*.

Let *ABC* be a circle, and *A* and *B* any two points in the circumference: the straight line drawn from *A* to *B* shall fall within the circle.
For if it do not, let it fall, if possible, without, as *AEB*.

Find *D* the centre of the circle *ABC*; [III. 1.

and join *DA*, *DB*; in the arc *AB* take any point *F*, join *DF*, and produce it to meet the straight line *AB* at *E*.

Then, because *DA* is equal to *DB*, [I. *Definition* 15.

the angle *DAB* is equal to the angle *DBA*. [I. 5. And because *AE*, a side of the triangle *DAE*, is produced to *B*, the exterior angle *DEB* is greater than the interior opposite angle *DAE*. [I. 16.

But the angle *DAE* was shewn to be equal to the angle *DBE*;

therefore the angle *DEB* is greater than the angle *DBE*.

But the greater angle is subtended by the greater side; [1. 19.

therefore *DB* is greater than *DE*.

But *DB* is equal to *DF*; [I. *Definition* 15.

therefore *DF* is greater than *DE*, the less than the greater;

which is impossible.

Therefore the straight line drawn from *A* to *B* does not fall without the circle.

In the same manner it may be shewn that it does not fall on the circumference.

Therefore it falls within the circle.

Wherefore, *if any two points* &c. q.e.d.

*THEOREM*.

*If a straight line drawn through the centre of a circle, bisect a straight line in it which does not pass through the * *centre it shall cut it at right angles; and if it cut it at right angles it shall bisect it*.

Let *ABC* be a circle; and let *CD*, a straight line drawn through the centre, bisect any straight line *AB*, which does not pass through the centre, at the point *F*: *CD* shall cut *AB* at right angles.

Take *E* the centre of the circle; and join *EA*, *EB*.[III. 1.

Then, because *AF* is equal to *FB*, [*Hypothesis*.

and *FE* is common to the two triangles *AFB*, *BFE*;

the two sides *AF*, *FE* are equal to the two sides *BF*,*FE*, each to each;

and the base *EA* is equal to the base *EB*; [I. *Def*. 15.

therefore the angle *AFE* is equal to the angle *BFE*. [I. 8.

But when a straight line, standing on another straight line, makes the adjacent angles equal to one another, each of the angles is called a right angle; [I. *Definition* 10.

therefore each of the angles *AFE*, *BFE* is a right angle.

Therefore the straight line *CD*, drawn through the centre, bisecting another *AB* which does not pass through the centre, also cuts it at right angles.

But let *CD* cut *AB* at right angles: *CD* shall also bisect *AB*; that is, *AF* shall be equal to *FB*.

The same construction being made, because *EA*, *EB*, drawn from the centre, are equal to one another, [I. *Def*. 15.

the angle *EAF* is equal to the angle *EBF*. [I. 5.

And the right angle *AFE* is equal to the right angle *BFE*.

Therefore in the two triangles *EAF*, *EBF*, there are two angles in the one equal to two angles in the other, each to each;

and the side *EF*, which is opposite to one of the equal angles in each, is common to both;

therefore their other sides are equal; [I. 26.

therefore *AF* is equal to *FB*.

*if a straight line*&c. q.e.d.

*THEOREM*.

*If in a circle two straight lines cut one another, which do not both pass through the centre, they do not bisect one another*.

Let *ABCD* be a circle, and *AC*, *BD* two straight lines in it, which cut one another at the point *E*, and do not both pass through the centre: *AC*, *BD* shall not bisect one another.

If one of the straight lines pass through the centre it is plain that it cannot be bisected by the other which does not pass through the centre.

But if neither of them pass through the centre, if possible, let *AE* be equal to *EC*, and *BE* equal to *ED*.

Take *F* the centre of the circle [III. 1.,

and join *EF*.

Then, because *FE*, a straight line drawn through the centre, bisects another straight line *AC* which does not pass through the centre; [*Hypothesis*.*FE* cuts *AC* at right angles; ' [III. 3.

therefore the angle *FEA* is a right angle.

Again, because the straight line *FE* bisects the straight line *BD*, which does not pass through the centre, [*Hyp*.*FE* cuts *BD* at right angles; [*III*. 3.

therefore the angle *FEB* is a right angle.

But the angle *FEA* was shewn to be a right angle;

therefore the angle *FEA* is equal to the angle *FEB*,[*Ax*.11.

the less to the greater; which is impossible.

Therefore *AC*, *BD* do not bisect each other.,

Wherefore, *if in a circle* &c. q.e.d.

*THEOREM*.

*If two circles cut one another, they shall not have the same centre*.

Let the two circles *ABC*, *CDG* cut one another at the
points *B*, *C*: they shall not have the same centre.

For, if it be possible, let *E* be their centre; join *EC*, and draw any straight line *EFG* meeting the circumferences at *F* and *G*.

Then, because *E* is the centre of the circle *ABC*, *EC* is equal to *EF*. [I. *Definition* 15.

Again, because *E* is the centre of the circle *CDG*, *EC* is equal to *EG*. [I. *Definition* 15.

But *EC* was shewn to be equal to *EF*;

therefore *EF* is equal to *EG*,[*Axiom* 1.

the less to the greater; which is impossible.

therefore *E* is not the centre of the circles *ABC*, *CDG*.

Wherefore, *if two circles* &c. q.e.d.

*THEOREM*.

*If two circles touch one another internally they shall not have the same centre*.

Let the two circles *ABC*, *CDE* touch one another internally at the point *G*: they shall not have the same centre.
For, if it be possible, let *F* be their centre; join *FC*, and draw any straight line *FEB*, meeting the circumferences at *E* and *B*.

Then, because *F* is the centre of the circle *ABC*, *FC* is equal to *FB*. [I. *Def*. 15.

Again, because *F* is the centre of the circle *CDE*,*FC* is equal to *FE*. [I. *Definition* 15.

But *FC* was shewn to be equal to *FB*

therefore *FE* is equal to *FB*,

the less to the greater; which is impossible.

Therefore *F* is not the centre of the circles *ABC*, *CDE*.

*if two circles' &c. q.e.d. *

*PROPOSITION 7.*THEOREM

*.*

*If any point he taken in the diameter of a circle which is not the centre, of all the straight lines which can be drawn from this point to the circumference, the greatest is that in which the centre is, and the other part of the diameter is the least; and, of any others, that which is nearer to the straight line which passes through the centre, is always greater than one more remote; and from the same point there can he drawn to the circumference two straight lines, and only two, which are equal to one another, one on each side of the shortest line*.

Let *ABCD* be a circle and *AD* its diameter, in which let any point *F* be taken which is not the centre; let *E* be the centre: of all the straight lines *FB*, *FC*, *FG*, &c. that can be drawn from *F* to the circumference, *FA*, which passes through *E*, shall be the greatest, and *FD*, the other part of the diameter *AD*, shall be the least; and of the others *FB* shall be greater than *FC*, and *FC* than *FG*.

Join *BE*, *CE*, *GE*.

Then, because any two sides of a triangle are greater than the third side, [I. 20.

therefore *BE*, *EF* are greater than *BF*.

But *BE* is equal to *AE*; [I.*Def*. 15.

therefore *AE*, *EF* are greater than *BF*,

that is, *AF* is greater than *BF*.

Again, because *BE* is equal to *CE*, [I. *Definition* 15.

and *EF* is common to the two triangles *BEF*, *CEF*;

the two sides *BE*, *EF* are equal to the two sides *CE*, *EF', each to each;*
but the angle

*BEF*is greater than the angle

*CEF*;

therefore the base

*FB*is greater than the base

*FC*. [I. 24.

In the same manner it may be shewn that

*FC*is greater than

*FG*.

*GF*,

*FE*are greater than

*EG*, [I. 20.

and that *EG* is equal to *ED*; [I. *Definition* 15.

therefore *GF*, *FE* are greater than *ED*.

Take away the common part *FE*, and the remainder *GF* is greater than the remainder *FD*.

Therefore *FA* is the greatest, and *FD* the least of all the straight lines from *F* to the circumference; and *FB* is greater than *FC*, and *FC* than *FG*.

Also, there can be drawn two equal straight lines from the point *F* to the circumference, one on each side of the shortest line *FD*.

For, at the point *E*, in the straight line *EF*, make the angle *FEH* equal to the angle *FEG*, [I. 23.

and join *FH*.

Then, because *EG* is equal to *EH*, [I. *Definition* 15.

and *EF* is common to the two triangles *GEF*, *HEF*;

the two sides *EG*, *EF* are equal to the two sides *EH*, *EF*, each to each;

and the angle *GEF* is equal to the angle *HEF*; [*Constr*.

therefore the base *FG* is equal to the base *FH*. [I. 4.

But, besides *FH*, no other straight line can be drawn from *F* to the circumference, equal to *FG*.

For, if it be possible, let *FK* be equal to *FG*.

Then, because *FK* is equal to *FG*, [*Hypothesis*.

and *FH* is also equal to *FG*,

therefore *FH* is equal to *FK*; [*Axiom* 1.

that is, a line nearer to that which passes through the centre is equal to a line which is more remote;

which is impossible by what has been already shewn.

Wherefore, *if any point be taken* &c. q.e.d.

PROPOSITION 8. *THEOREM*.

*If any point be taken without a circle, and straight lines be drawn from it to the circumference, one of which passes through the centre; of those which fall on the concave circumference, the greatest is that which passes through the centre, and of the rest, that which is nearer to the one passing through the centre is always greater than one more remote; but of those which fall on the* *convex circumference, the least is that between the point without the circle and the diameter; and of the rest, that which is nearer to the least is always less than one more remote; and from the same point there can he drawn to the circumference two straight lines, and only two, which are equal to one another, one on each side of the shortest line*.

Let *ABC* be a circle, and *D* any point without it, and from *D* let the straight lines *DA*, *DE*, *DF*, *DC* be drawn to the circumference, of which *DA* passes through the centre: of those which fall on the concave circumference *AEFC*, the greatest shall be *DA* which passes through the centre, and the nearer to it shall be greater than the more remote, namely, *DE* greater than *DF*, and *DF* greater than *DC*;

but of those which fall on the convex circumference *GKLH*, the least shall be *DG* between the point *D* and the diameter *AG*, and the nearer to it shall be less than the more remote, namely, *DK* less than *DL*, and *DL* less than *DH*.

Take *M*, the centre of the circle *ABC*, [III. 1.

and join *ME*, *MF*, *MC*, *MH*, *ML*, *MK*.

Then, because any two sides of a triangle are greater than the third side, [I. 20.

therefore *EM*, *MD* are greater than *ED*.

But *EM* is equal to *AM*;[I.*Def*.15. therefore *AM*, *MD* are greater than *ED*, that is, *AD* is greater than *ED*.

Again, because *EM* is equal to *FM*,

and *MD* is common to the two triangles *EMD*, *FMD*;

the two sides *EM*, *MD* are equal to the two sides *FM*, *MD*, each to each;

but the angle *EMD* is greater than the angle *FMD*;

therefore the base *ED* is greater than the base *FD*. [1. 24.

In the same manner it may be shewn that *FD* is greater than *CD*.

Therefore *DA* is the greatest, and *DE* greater than *DF* and *DF* greater than *DC*.

Again, because *MK*, *KD* are greater than *MD*, [I, 20.

and *MK* is equal to *MG* [I. *Definition* 15.

the remainder *KD* is greater than the remainder *GD*,

that is, *GD* is less than *KD*.

And because *MLD* is a triangle, and from the points
*M*, *D*, the extremities of its side *MD* the straight lines
*MK*, *DK* are drawn to the point *K* within the triangle,
therefore *MK*, *KD* are less than *ML*, *LD*; [I. 21.

and *MK* is, equal to *ML* ; [I. *Definition* 15.

therefore the remainder *KD* is less than the remainder *LD*.

In the same manner it may be shewn that *LD* is less
than *HD*.

Therefore *DG* is the least, and *DK* less than *DL*, and *DL*
less than *DH*.

Also, there can be drawn two equal straight lines from
the point *D* to the circumference, one on each side of the
least line.

For, at the point *M*, in the straight line *MD*, make the
angle *DMB* equal to the angle *DMK*, [I. 23.

and join *DB*.

Then, because *MK* is equal to *MB*,
and *MD* is common to the two triangles *KMD*, *BMD* ;

the two sides *KM*, *MD* are equal to the two sides *BM*, *MD*,
each to each ;

and the angle *DMK* is equal to the angle *DMB* ; [*Constr*.

therefore the base *DK* is equal to the base *DB*. [I. 4.

But, besides *DB*, no other straight line can be drawn
from *D* to the circumference, equal to *DK*.

For, if it be possible, let *DN* be equal to *DK*.

Then, because *DN* is equal to *DK*,
and *DB* is also equal to *DK*,

therefore *DB* is equal to *DN*; [*Axiom* 1.

that is, a line nearer to the least is equal to one which is more remote ; which is impossible by what has been already shewn.

Wherefore,*if any point be taken*&c. q.e.d.

*THEOREM*.

*If a point be taken within a circle, from which there fall more than two equal straight lines to the circumference, that point is the centre of the circle*.

Let the point *D* be taken within the circle *ABC*, from which to the circumference there fall more than two equal straight lines, namely *DA*, *DB*, *DC*: the point *D* shall be the centre of the circle.

For, if not, let *E* be the centre; join 'DE* and produce it both ways to meet the circumference at *F* and *G*; then *FG* is a diameter of the circle.*

Then, because in *FG*, a diameter of the circle *ABC*, the point *D' is taken, which is not the centre, *DG* is the greatest straight line from *D* to the circumference, and *DC* is greater than *DB*, and *DB* greater than *DA*; [III. 7.*
but they are likewise equal, by hypothesis;

which is impossible.

Therefore

*E*is not the centre of the circle

*ABC*.

In the same manner it may be shewn that any other point than *D* is not the centre;

therefore *D* is the centre of the circle *ABC*.

Wherefore, *if a point be taken* &c. q.e.d.

*THEOREM*.

*One circumference of a circle cannot cut another at more than two points*.

If it be possible, let the circumference *ABC* cut the circumference *DEF* at more than two points, namely, at the points *B*, *G*, *F*.

Take *K*, the centre of the circle *ABC*, [II. 1.

and join *KB*, *KG*, *KF*.

Then, because *K* is the centre of the circle *ABC*,

therefore *KB*,*KG*, *KF* all equal to each other. [I. *Def*. 15.

And because within the circle *DEF*, the point *K* is taken, from which to the circumference *DEF* fall more than two equal straight lines *KB*, *KG*, *KF*, therefore *K* is the centre of the circle *DEF*. [III. 9.

But *K* is also the centre of the circle *ABC*. [*Construction*.

Therefore the same point is the centre of two circles which cut one another;

which is impossible. [III. 5.

Wherefore, *one circumference* &c. q.e.d.

*THEOREM*.

*If two circles touch one another internally, the straight line which joins their centres, being produced, shall pass through the point of contact*.

Let the two circles *ABC*, *ADE* touch one another internally at the point *A*; and let *F* be the centre of the circle *ABC*, and *G* the centre of the circle *ADE*: the straight line which joins the centres *F*, *G*, being produced, shall pass through the point *A* .

For, if not, let it pass otherwise, if possible, as *FGDH*, and join *AF*,*AG*.

Then, because *AG*, *GF* are greater than *AF*, [I. 20.

and *AF* is equal to *HF*, [I. *Def*. 15. therefore *AG*, *GF*, are greater than *HF*. Take away the common part *GF*; therefore the remainder *AG* is greater than the remainder *HG*. But *AG* os equal to *DG*. [I. *Definition* 15.

Therefore *DG* is, greater than *HG*, the less than the greater;

which is impossible.

Therefore the straight line which joins the points *F*, *G*, being produced, cannot pass otherwise than through the point *A*, that is, it must pass through *A*.

*if two circles*&c. q.e.d.

*THEOREM*.

*If two circles touch one another externally, the straight line which joins their centres shall pass through the point of contact*.

Let the two circles *ABC*, *ADE* touch one another externally at the point *A*; and let *F* be the centre of the circle *ABC*, and *G* the centre of the circle *ADE*: the straight line which joins the points *F*, *G*, shall pass through the point *A* .

For, if not, let it pass otherwise, if possible, as *FCDG*, and join *FA*, *AG*.

Then, because *F* is the centre of the circle *ABC*, *FA* is equal to *FC*; [I.*Def*.16.

and because *G* is the centre of the circle *ADE*, *GA* is equal to *GD*; therefore *FA*, *AG* are equal to *FC*, *DG*. [*Axiom 2*.

Therefore the whole *FG* is greater than *FA*, *AG*. But *FG* is also less than *FA*, *AG*; [I. 20.

which is impossible.

Therefore the straight line which joins the points *F*, *G*, cannot pass otherwise than through the point *A*, that is, it must pass through *A*.

Wherefore, *if two circles* &c. q.e.d.

*THEOREM*.

*One circle cannot touch another at more points than one, whether it touches it on the inside or outside*.

For, if it be possible, let the circle *EBF* touch the circle *ABC* at more points than one; and first on the inside, at the points *B*, *D*. Join *BD*, and draw *GH* bisecting *BD* at right angles. [I. 10, 11.

Then, because the two points *B*, *D* are in the circumference of each of the circles, the straight line *BD* falls within each of them; [III. 2.

and therefore the centre of each circle is in the straight line *GH* which bisects *BD* at right angles; [III. 1, *Corol*.

therefore *GH* passes through the point of contact. [III. 11.

But *GH* does not pass through the point of contact, because the points *B*, *D* are out of the line *GH*;

which is absurd.

Therefore one circle cannot touch another on the inside at more points than one.

Nor can one circle touch another on the outside at more points than one.

For, if it be possible, let the circle *ACK* touch the circle *ABG* at the points *A*, *C*. Join *AC*.

Then, because the two points *A*, *C* are in the circumference of the circle *ACK*, the straight line *AC* which joins them, falls within the circle *ACK*; [III. 2.

but the circle *ACK* is without the circle *ABC*; [*Hypothesis*.

therefore the straight line *AC* is without the circle *ABC*.

But because the two points *A*,*C* are in the circumference of the circle *ABC*, the straight line *AC* falls within the circle *ABC*; [III. 2.

which is absurd.

Therefore one circle cannot touch another on the outside at more points than one.

And it has been shewn that one circle cannot touch another on the inside at more points than one.

Wherefore,*one circle*&c. q.e.d.

*THEOREM*.

*Equal straight lines in a circle are equally distant from the centre: and those which are equally distant from the centre are equal to one another*.

Let the straight lines *AB*, *CD* in the circle *ABDC*, be equal to one another: they shall be equally distant from the centre.

Take *E*, the centre of the circle *ABDC*; [III. 1.

and from *E* draw *EF*, *EG* perpendiculars to *AB*, *CD*; [I. 12.

and join *EA*,*EC*.

Then, because the straight line *EF*, passing through the centre, cuts the straight line *AB*, which does not pass through the centre, at right angles, it also bisects it; [III. 3.

therefore *AF* is equal to *FB*, and *AB* is double of *AF*.

For the like reason *CD* is double of *CG*.

But *AB* is equal to *CD*; [*Hypothesis*.

therefore *AF* is equal to *CG*. [*Axiom* 7.

And because *AE* is equal to *CE*, [I. *Definition* 15. the square on *AE* is equal to the square on *CE*.

But the squares on *AF*, *FE* are equal to the square on *AE*, because the angle *AFE* is a right angle; [I. 47.

and for the like reason the squares on *CG*, *GE* are equal to the square on *CE*;

therefore the squares on *AF*, *FE* are equal to the squares on *CG*, *GE*. [*Axiom* 1.

But the square on *AF* is equal to the square on *CG*, because *AF* is equal to *CG*;

therefore the remaining square on *FE* is equal to the remaining square on *GE*; [*Axiom* 3.

and therefore the straight line *EF* is equal to the straight line *EG*.

But straight lines in a circle are said to be equally distant from the centre, when the perpendiculars drawn to them from the centre are equal; [III. *Definition* 4.

therefore *AB*, *CD* are equally distant from the centre.

Next, let the straight lines *AB*, *CD* be equally distant from the centre, that is, let *EF* be equal to *EG*, *AB* shall be equal to *CD*.

For, the same construction being made, it may be shewn, as before, that *AB* is double of *AF*, and *CD* double of *CG* and that the squares on *EF*,*FA* are equal to the squares on *EG*,*GC*; but the square on *EF* is equal to the square on *EG*, because *EF* is equal to *EG*; [*Hypothesis*. therefore the remaining square on *FA* is equal to the remaining square on *GC*, [*Axiom* 3.

and therefore the straight line *AF* is equal to the straight line *CG*.

But *AB* was shewn to be double of *AF*, and *CD* double of *CG*. Therefore *AB* is equal to *CD*. [*Axiom* 6.

Wherefore, *equal straight lines* &c. q.e.d.

*THEOREM*.

The diameter is the greatest straight line in a circle; and, of all others, that which is nearer to the centre is cdways greater than one more remote; and the greater is nearer to the centre than the less.

Let *ABCD* be a circle, of which *AD* is a diameter, and *E* the centre; and let *BC* be nearer to the centre than FG:*AD* shall be greater than any straight line *BC* which is not a diameter, and *BC* shall be greater than *FG*.

From the centre *E* draw *EH*, *EK* perpendiculars to *BC*, *FG*, [I. 12.

and join *EB*, *EC*, *EF*.

Then, because *AE* is equal to *BE*, and *ED* to *EC*, [I.*Def*.15.

therefore *AD* is equal to *BE*, *EC*; [*Axiom* 2.

but *BE*, *EC* are greater than *BC*; [I. 20.

therefore also *AD* is greater than *BC*.

And, because *BC* is nearer to the centre than *FG*, [*Hypothesis*.*EH* is less than *EK*. [III.Def.5.

Now it may be shewn, as in the preceding proposition, that *BC* is double of *BH*, and *FG* double of *FK*, and that the squares on *EH*, *HB* are equal to the squares on *EK',' *KF*.*
But the square on

*EH*is less than the square on

*EK*, because

*EH*is less than

*EK*;

therefore the square on

*HB*is greater than the square on

*KF*;

and therefore the straight line

*BH*is greater than the straight line

*FK*;

and therefore

*BC*is greater than

*FG*.

Next, let *BC* be greater than *FG*: *BC* shall be nearer to the centre than *FG*, that is, the same construction being made, *EH* shall be less than *EK*.

For, because BC is greater than FG, BH is greater than FK.

But the squares on BH, HE are equal to the squares on FK,KE;

and the square on *BH* is greater than the square on *FK*;

because *BH* is greater than *FK*;

therefore the square on *HE* is less than the square on *KE*;

and therefore the straight line *EH* is less than the straight line *EK*.

Wherefore, *the diameter* &c. q.e.d.

*THEOREM*.

*The straight line drawn at right angles to the diameter of a circle from the extremity of it, falls without the circle; and no straight line can be drawn from the extremity, between that straight line and the circumference, so as not to cut the circle*. Let *ABC* be a circle, of which *D* is the centre and *AB* a diameter: the straight line drawn at right angles to *AB*, from its extremity *A*, shall fall without the circle.

For, if not, let it fall, if possible, within the circle, as *AC*, and draw *DC* to the point *C*, where it meets the circmference.

Then, because *DA* is equal to *DC*, [I. *Definition* 15.

the angle *DAC* is equal to the angle *DCA*. [I. 5.

But the angle *DAC* is a right angle; [*Hypothesis*.

therefore the angle *DCA* is a right angle;

and therefore the angles *DAC*, *DCA* are equal to two right angles; which is impossible. [I. 17.

Therefore the straight line drawn from *A* at right angles to *AB* does not fall within the circle.

And in the same manner it may be shewn that it does not fall on the circumference.

Therefore it must fall without the circle, as *AE*.

Also between the straight line *AE* and the circumference, no straight line can be drawn from the point *A*, which does not cut the circle.

For, if possible, let *AF* be between them; and from the centre *D* draw *DG* perpendicular to *AF*; [I. 12.

let *DG* meet the circumference at *H*.

Then, because the angle *DGA* is a right angle, [*Construction*.

the angle *DAG* is less than a right angle; [I. 17.

therefore *DA* is greater than *DG*. [1.19.

But *DA* is equal to *DH*; [I. *Definition* 15.

therefore *DH* is greater than *DG*, the less than the greater;

which is impossible.

Therefore no straight line can be drawn from the point *A* between *AE* and the circumference, so as not to cut the circle. Wherefore, *the straight line* &c. q.e.d.

Corollary. From this it is manifest, that the straight line which is drawn at right angles to the diameter of a circle from the extremity of it, touches the circle; [III. *Def*. 2. and that it touches the circle at one point only,

because if it did meet the circle at two points it would fall within it. [III. 2.

Also it is evident, that there can be but one straight line which touches the circle at the same point.

*PROBLEM*.

*To draw a straight line from a given point, either without or in the circumference, which shall touch a given circle*.

First, let the given point *A* be without the given circle *BCD*: it is required to draw from *A* a straight line, which shall touch the given circle.

Take *E*, the centre of the circle, [III. 1.

and join *AE* cutting the circumference of the given circle at *D*;

and from the centre *E*, at the distance *EA*, describe the circle *AFG*; from the point *D* draw *DF* at right angles to *EA*,[I.11.

and join *EF* cutting the circumference of the given circle at *B*;

join *AB*. *AB* shall touch the circle *BCD*.

For, because *E* is the centre of the circle *AFG*, *EA* is equal to *EF*. [I. *Definition* 15. And because *E* is the centre of the circle *BCD*, *EB* is equal to *ED*. [I. *Definition* 15.

Therefore the two sides *AE*, *EB* are equal to the two sides *FE*, *ED*, each to each;

and the angle at *E* is common to the two triangles *AEB*, *FED*;

therefore the triangle *AEB* is equal to the triangle *FED*, and the other angles to the other angles, each to each, to which the equal sides are opposite; [I. 4.

therefore the angle *ABE* is equal to the angle *FDE*.

But the angle *FDE* is a right angle; [*Construction*.

therefore the angle *ABE* is a right angle. [*Axiom* 1.

And *EB* is drawn from the centre; but the straight line drawn at right angles to a diameter of a circle, from the extremity of it, touches the circle; [III. 16, *Corollary*.

therefore *AB* touches the circle.

And *AB* is drawn from the given point *A*. q.e.f.

But if the given point be in the circumference of the circle, as the point *D*, draw *DE* to the centre *E*, and *DF* at right angles to *DE*; then *DF* touches the circle. [III. 16, *Cor*.

*THEOREM*.

*If a straight line touch a circle the straight line drawn from the centre to the point of contact shall be perpendicular to the line touching the circle*.

Let the straight line *DE* touch the circle *ABC* at the point *C*; take *F*, the centre *of* the circle *ABC*, and draw the straight line *FC*: *FC* shall be perpendicular to *DE*.

For if not, let *FG* be drawn from the point *F* perpendicular to *DE*, meeting the circumference at *B*.

Then, because *FGC* is a right angle, [*Hypothesis*. *FCG* is an acute angle; [I. 17.

and the greater angle of every triangle is subtended by the greater side; [I. 19.

therefore *FC* is greater than *FG*.

But *FC* is equal to *FB*; [I. *Definition* 15.

therefore *FB* is greater than *FG*, the less than the greater;

which is impossible.

Therefore *FG* is not perpendicular to *DE*.

In the same manner it may be shewn that no other straight line from *F* is perpendicular to *DE*, but *FC*;

therefore *FC* is perpendicular to *DE*.

*if a straight line*&c. q.e.d.

*THEOREM*.

*If a straight line touch a circle, and from the point of contact a straight line he drawn at right angles to the touching line, the centre of the circle shall he in that line*.

Let the straight line *DE* touch the circle *ABC* at *C*, and from *C* let *CA* be drawn at right angles to *DE*: the centre of the circle shall be in *CA*.

For, if not, if possible, let *F* be the centre, and join *CF*.

Then, because *DE* touches the circle *ABC*, and *FC* is drawn from the centre to the point of contact, *FC* is perpendicular to *DE*; [III. 18.

therefore the angle *FCE* is a right angle.

But the angle *ACE* is also a right angle; [*Construction*.;

therefore the angle *FCE* is equal to the angle *ACE*, [*Ax*.11.

the less to the greater; which is impossible. Therefore *F* is not the centre of the circle *ABC*.

In the same manner it may be shewn that no other point out of *CA* is the centre; therefore the centre is in *CA*.

Wherefore, *if a straight line* &c. q.e.d.

*THEOREM*.

*The angle at the centre of a circle is double of the angle at the circumference on the same base, that is, on the same arc*.

Let *ABC* be a circle, and *BEC* an angle at the centre, and *BAC* an angle at the circumference, which have the same arc, *BC*, for their base: the angle *BEC* shall be double of the angle *BAC*.

Join *AE*, and produce it to *F*.

First let the centre of the circle be within the angle *BAC*.

Then, because *EA* is equal to *EB*, the angle *EAB* is equal to the angle *EBA*; [I. 5.

therefore the angles *EAB*, *EBA* are double of the angle *EAB*.

But the angle *BEF* is equal to the angles *EAB*, *EBA*; [1. 32. therefore the angle *BEF* is double of the angle *EAB*.

For the same reason the angle *FEC* is double of the angle *BAC*.

Therefore the whole angle *BEC* is double of the whole angle *BAC*.

Next, let the centre of the circle be without the angle *BAC*.

Then it maybe shewn, as inthe first case, that the angle *FEC* is double of the angle *FAC*, and that the angle *FEB*, a part of the first, is double of the angle *FAB*, a part of the other;

therefore the remaining angle BEC is double of the remaining angle BAC.

Wherefore, *the angle at the centre* &c. q.e.d.

*THEOREM*.

The angles in the same segment of a circle are equal to one another.

Let ABGD be a circle, and BAD, BED angles in the same segment BAED: the angles BAD, BED shall be equal to one another. Take F the centre of the circle ABGD. [III. 1.

First let the segment *BAED* be greater than a semicircle.

Join *BF*, *DF*.

Then, because the angle *BFD* is at the centre, and the angle *BAD* is at the circumference, and that they have the same arc for their base, namely, *BCD*;

therefore the angle *BFD* is double of the angle *BAD*.[III.20.

For the same reason the angle *BFD* is double of the angle *BED*.

Therefore the angle *BAD* is equal to the angle *BED*. [*Ax*. 7.

Next, let the segment *BAED* be not greater than a semicircle.

Draw *AF* to the centre, and produce it to meet the circumference at *C*, and join *CE*.

Then the segment *BAEC* is greater than a semicircle, and therefore the angles *BAC*, *BEC* in it, are equal, by the first case. For the same reason, because the segment *CAED* is greater than a semicircle, the angles *CAD*, *CED* are equal.

Therefore the whole angle *BAD* is equal to the whole angle *BED*. [*Axiom* 2.

Wherefore, *the angles in the same segment* &c. q.e.d.

*THEOREM*.

The opposite angles of any quadrilateral figure inscribed in a circle are together equal to two right angles.

Let *ABCD* be a quadrilateral figure inscribed in the circle *ABCD*: any two of its opposite angles shall be together equal to two right angles.

Join *AC*, *BD*.

Then, because the three angles of every triangle are together equal to two right angles, [I. 32.

the three angles of the triangle *CAB*, namely, *CAB*, *ACB*, *ABC* are together equal to two right angles.

But the angle *CAB* is equal to the angle *CDB*, because they are in the same segment *CDAB*; [III. 21.

and the angle *ACB* is equal to the angle *ADB*, because they are in the same segment *ADCB*;

therefore the two angles *CAB*, *ACB* are together equal to the whole angle *ADC*. [*Axiom* 2.

To each of these equals add the angle *ABC*;

therefore the three angles *CAB*, *ACB*, *ABC*, are equal to the two angles *ABC*, *ADC*.

But the angles *CAB*, *ACB*, *ABC* are together equal to two right angles; [I. 32.

therefore also the angles *ABC*, *ADC* are together equal to two right angles.

In the same manner it may be shewn that the angles *BAD*, *BCD* are together equal to two right angles.

Wherefore, *the opposite angles* &c. q.e.d.

*THEOREM*.

*On the same straight line, and on the same side of it, there cannot he two similar segments of circles, not coinciding with one another*.

If it be possible, on the same straight line *AB*, and on the same side of it, let there be two similar segments of circles *ACB*, *ADB*, not coinciding with one another.

Then, because the circle *ACB* cuts the circle *ADB* at the two points *A*, *B*, they cannot cut one another at any other point; [III. 10.

therefore one of the segments must fall within the other; let *ACB* fall within *ADB*; draw the straight line *BCD*, and join *AC*, *AD*.

Then, because *ACB*, *ADB* are, by hypothesis, similar segments of circles, and that similar segments of circles contain equal angles, [III. *Definition* 11.

therefore the angle *ACB* is equal to the angle *ADB*;

that is, the exterior angle of the triangle *ACD* is equal to the interior and opposite angle;

which is impossible. [I. 16.

*on the same straight line*&c. q.e.d.

*THEOREM*.

*Similar segments of circles on equal straight lines are equal to one another*.
Let *AEB*, *CFD* be similar segments of circles on the equal straight lines *AB*, *CD*: the segment *AEB* shall be equal to the segment *CFD*.

For if the segment *AEB* be applied to the segment *CFD*, so that the point *A* may be on the point *C*, and the straight line *AB* on the straight line *CD*, the point *B* will coincide with the point *D*, because *AB* is equal to *CD*.

Therefore, the straight line *AB* coinciding with the straight line *CD*, the segment *AEB* must coincide with the segment *CFD*; [III. 23.

and is therefore equal to it.

Wherefore, *similar segments* &c. q.e.d.

*PROBLEM*.

*A segment of a circle being given, to describe the circle of which it is a segment*.

Let *ABC* be the given segment of a circle: it is required to describe the circle of which it is a segment.

Bisect *AC* at *D*; [I. 10.

from the point *D* draw *DB* at right angles to *AC*; [1. 11.

and join *AB*.

First, let the angles *ABD*, *BAD*, be equal to one another.

Then *DB* is equal to *DA*; [I. 6. but *DA* is equal to *DC*; [*Construction*.

therefore *DB* is equal to *DC*. [*Axiom* 1.

Therefore the three straight lines *DA*,*DB*,*DC* are all equal;

and therefore *D* is the centre of the circle. [III. 9.

From the centre *D*, at the distance of any of the three
*DA*, *DB*, *DC*, describe a circle; this will pass through
the other points, and the circle of which *ABC* is a segment
is described.

And because the centre *D* is in *AC*, the segment *ABC*
is a semicircle.

Next, let the angles *ABD*, *BAD* be not equal to one
another.

At the point *A*, in the straight line *AB*, make the angle
*BAE* equal to the angle *ABD* ; [I, 23.

produce *BD*, if necessary, to meet *AE* at *E*, and join *EC*.

Then, because the angle *BAE* is equal to the angle
*ABE*, [*Construction*.*EA* is equal to *EB*. [I. 6.

And because *AD* is equal to *CD*, [*Construction*.

and *DE* is common to the two triangles *ADE*, *CDE* ;

the two sides *AD*, *DE* are equal to the two sides *CD*, *DE*,
each to each ;

and the angle *ADE* is equal to the angle *CDE*, for each of
them is a right angle ; [*Construction*.

therefore the base *EA* is equal to the base *EC*. [I. 4.

But *EA* was shewn to be equal to *EB* ;

therefore *EB* is equal to *EC*. [*Axiom* 1.

Therefore the three straight lines *EA*, *EB*, *EC* are all equal ;

and therefore *E* is the centre of the circle. [III. 9.

From the centre *E*, at the distance of any of the three
*EA*, *EB*, *EC*, describe a circle ; this will pass through the
other points, and the circle of which *ABC* is a segment is
described.

And it is evident, that if the angle *ABD* be greater
than the angle *BAD*, the centre *E* falls without the seg-
ment *ABC*, which is therefore less than a semicircle ; but
if the angle *ABD* be less than the angle *BAD*, the centre
*E* falls within the segment *ABC*, which is therefore greater
than a semicircle.

Wherefore, *a segment of a circle being given, the circle *

*. q.e.f. *

*PROPOSITION 26.*THEOREM

*.*

*In equal circles, equal angles stand on equal arcs, whether they be at the centres or circumferences*.

Let *ABC*, *DEF* be equal circles; and let *BGC*, *EHF* be equal angles in them at their centres, and *BAC*, *EDF* equal angles at their circumferences: the arc *BKC* shall be equal to the arc *ELF*.

Join *BC*, *EF*.

Then, because the circles *ABC*, *DEF* are equal, [*Hyp*.

the straight lines from their centres arc equal; [III. *Def*. 1.

therefore the two sides *BG*, *GC* are equal to the two sides *EH*, *HF*, each to each;

and the angle at *G* is equal to the angle at *H*; [*Hypothesis*.

therefore the base *BC* is equal to the base *EF*. [I. 4.

And because the angle at *A* is equal to the angle at *D,[*Hyp*.*
the segment

*BAC*dissimilar to the segment

*EDF*; [III.

*Def*. ll.

and they are on equal straight lines

*BC*,

*EF*.

But similar segments of circles on equal straight lines are equal to one another; [III. 24.

therefore the segment

*BAC*be equal to the segment

*EDF*.

But the whole circle *ABC* is equal to the whole circle *DEF*; [*Hypothesis*.

therefore the remaining segment *BKC* is equal to the remaining segment ELF; [Axiom 3.

therefore the arc *BKC* is equal to the arc *ELF*.

*in equal circies*&c. q.e.d.

*THEOREM*.

*In equal circles, the angles which stand on equal arcs are equal to one another whether they he at the centres or circumferences*.

Let *ABC*, *DEF* be equal circles, and let the angles *BGC*, *EHF* at their centres, and the angles *BAC*, *EDF* at their circumferences, stand on equal arcs *BC*, *EF*: the angle *BGC* shall be equal to the angle *EHF*, and the angle *BAC* equal to the angle *EDF*.

If the angle *BGC* be equal to the angle *EHF*, it is manifest that the angle *BAC* is also equal to the angle *EDF*. [III. 20, *Axiom* 7.

But, if not, one of them must be the greater. Let *BGC* be the greater, and at the point *G*, in the straight line *BG*, make the angle *BGK* equal to the angle *EHF*. [I. 23.

Then, because the angle *BGK* is equal to the angle *EHF*, and that in equal circles equal angles stand on equal arcs, when they are at the centres, [III. 26.

therefore the arc *BK* is equal to the arc *EF*.

But the arc *EF* is equal to the arc *BC*; [*Hypothesis*.

therefore the arc *BK* is equal to the arc *BC*, [*Axiom* 1.

the less to the greater; which is impossible.

Therefore the angle *BGC* is not unequal to the angle *EHF*, that is, it is equal to it.

And the angle at *A* is half of the angle *BGC*, and the angle at *D* is half of the angle *EHF*; [III. 20.

therefore the angle at A is equal to the angle at *D*. [*Ax*. 7.

*THEOREM*.

*In equal circles, equal straight lines cut off equal arcs, the greater equal to the greater, and the less equal to the less*.

Let *ABC*, *DEF* be equal circles, and *BC*,*EF* equal straight lines in them, which cut off the two greater arcs *BAC*, *EDF*, and the two less arcs *BGC*, *EHF* the greater arc *BAC* shall be equal to the greater arc *EDF* and the less arc *BGG* equal to the less arc *EHF*.

Take *K*, *L*, the centres of the circles, [III. 1.

and join *BK*, *KC*, *EL*, *LF*.

Then, because the circles are equal, [*Hypothesis*. the straight lines from their centres are equal; [III. *Def*. 1. therefore the two sides *BK*, *KC* are equal to the two sides *EL*, *LF*, each to each;

and the base *BC* is equal to the base *EF*; [*Hypothesis*.

therefore the angle *BKC* is equal to the angle *ELF*. [I. 8.

But in equal circles equal angles stand on equal arcs, when they are at the centres, [III. 26.

therefore the arc *BGG* is equal to the arc *EHF*.

But the circumference *ABGC* is equal to the circumference *DEHF*; [*Hypothesis*. therefore the remaining arc *BAC* is equal to the remaining arc *EDF*. [*Axiom* 3.

*in equal circles*&c. q e.d.

*THEOREM*.

*In equal circles, equal arcs are subdtended by equal straight lines*.

Let *ABC*, *DEF* be equal circles, and let *BGG*, *EHF* be equal arcs in them, and join *BC*, *EF*: the straight line *BC* shall be equal to the straight line *EF*.

Take *K*, *L*, the centres of the circles, [III. 1.

and join *BK*, *KG*, *EL*, *LF*.

Then, because the arc *BCG* is equal to the arc *EHF*, [*Hypothesis*.

the angle *BKC* is equal to the angle *ELF*. [III. 27.

And because the circles *ABC*, *DEF* are equal, [*Hypothesis*.

the straight lines from their centres are equal; [III. *Def*. 1.

therefore the two sides *BK*, *KC* are equal to the two sides *EL*, *LF*, each to each;

and they contain equal angles;

therefore the base *BC* is equal to the base *EF*. [I. 4.

Wherefore, *in equal circles* &c. q.e.d.

*PROBLEM*.

*To bisect a given arc, that is, to divide it into two equal parts*. Let *ADB* be the given arc: it is required to bisect it.

Join AB;

bisect it at *C*; [I. 10.

from the point *C* draw *CD* at right angles to *AB* meeting the arc at *D*. [I. 11

The arc *ADB* shall be bisected at the point *D*.

Join *AD*, *DB*.

Then, because *AC* is equal to *CB*, [*Construction*.

and *CD* is common to the two triangles *ACD*, *BCD*;

the two sides *AC*, *CD* are equal to the two sides *BC*, *CD*, each to each;

and the angle *ACD* is equal to the angle *BCD*, because each of them is a right angle; [*Construction*.

therefore the base *AD* is equal to the base *BD*. [I. 4.

But equal straight lines cut off equal arcs, the greater equal to the greater, and the less equal to the less; [III. 28.

and each of the arcs *AD*, *DB* is less than a semi-circumference, because *DC*, if produced, is a diameter; [III. 1. *Cor*. therefore the arc *AD* is equal to the arc *DB*.

Wherefore *the given arc is bisected at D*. q.e.f.

*THEOREM*.

*In a circle the angle in a semicircle is a right angle; but the angle in a segment greater than a semicircle is less than a right angle; and the angle in a segment less than a semicircle is greater than a right angle*.

Let *ABCD* be a circle, of which *BC* is a diameter and *E* the centre; and draw *CA*, dividing the circle into the segments *ABC*, *ADC*, and join *BA*, *AD*, *DC*: the angle in the semicircle *BAC* shall be a right angle; but the angle in the segment *ABC*, which is greater than a semicircle, shall be less than a right angle; and the angle in the segment *ADC'[', which is less than a semicircle, shall be greater than a right angle.*

Join *AE*, and produce *BA* to *F*.

Then, because *EA* is equal to *EB*, [I. Definition 15.

the angle *EAB* is equal to the angle *EBA*; [1.5.

and, because *EA* is equal to *EC*, the angle *EAC* is equal to the angle *ECA*; therefore the whole angle *BAC* is. equal to the two angles, *ABC*, *ACB*. [*Axiom* 2.

But *FAC*, the exterior angle of the triangle *ABC*, is equal to the two angles *ABC*, *ACB*; [I. 32.

therefore the angle *BAC* is equal to the angle *FAC*, [*Ax*. I.

and therefore each of them is a right angle. [I. *Def*. 10.

Therefore the angle in a semicircle *BAC* is a right angle.

And because the two angles *ABC*, *BAC*, of the triangle ABG, are together less than two right angles, [I. 17.

and that *BAC* has been shewn to be a right angle, therefore the angle *ABC* less than a right angle.

Therefore the angle in a segment *ABC*, greater than a semicircle, is less than a right angle.

And because *ABCD* is a quadrilateral figure in a circle, any two of its opposite angles are together equal to two right angles; [III. 22.

therefore the angles *ABC*, *ADC* are together equal to two right angles.

But the angle *ABC* has been shewn to be less than a right angle;

therefore the angle *ADC* is greater than a right angle.

Therefore the angle in a segment *ADC*, less than a semicircle, is greater than a right angle.

Wherefore, *the* angle &c. q.e.d. Corollary. From the demonstration it is manifest that if one angle of a triangle bo equal to the other two, it is a right angle.

For the angle adjacent to it is equal to the same two angles; [I. 32.

and when the adjacent angles are equal, they are right angles. [I. *Definition* 10.

*THEOREM*.

*If a straight line touch a circle, and froam the point of contact a straight line he drawn cutting the circle, the angles which this line makes with the line touching the circle shall be equal to the angles which are in the alternate segments of the circle*.

Let the straight line *EF* touch the circle *ABCD* at the point *B*, and from the point *B* let the straight line *BD* be drawn, cutting the circle: the angles which *BD* makes with the touching line *EF*, shall be equal to the angles in the alternate segments of the circle; that is, the angle *DBF* shall be equal to the angle in the segment *BAD*, and the angle *DBE* shall be equal to the angle in the segment *BCD*

From the point *B* draw *BA* at right angles to *EF*, [1. 11.

and take any point *C* in the, arc *BD*, and join *AD*, *DC*, *CB*.

Then, because the straight line *EF* touches the circle *ABCD* at the point *B*, [*Hyp*. and *BA* is drawn at right angles to the touching line from the point of contact *B*, [*Construction*.

therefore the centre of the circle is in *BA*. [III. 19.

Therefore the angle *ADB*, being in a semicircle, is a right angle. [III. 31.

Therefore the other two angles *BAD*, *ABD* are equal to a right angle. [I. 32.

But *ABF* is also a right angle. [*Construction*. Therefore the angle *ABF* is equal to the angles *BAD*, *ABD*.

From each of these equals take away the common angle *ABD*;

therefore the remaining angle *DBF* is equal to the remaining angle *BAD*, [*Axiom* 3.

which is in the alternate segment of the circle.

And because *ABCD* is a quadrilateral figure in a circle, the opposite angles *BAD*, *BCD* are together equal to two right angles. [III. 22.

But the angles *DBF*, *DBE* are together equal to two right angles. [I. 13.

Therefore the angles *DBF*, *DBE* are together equal to the angles *BAD*, *BCD*.

And the angle *DBF* has been shewn equal to the angle *BAD*;

therefore the remaining angle *DBE* is equal to the remaining angle *BCD*, [*Axiom* 3.

which is in the alternate segment of the circle.

Wherefore, *if a straight line* &c. q.e.d.

*PROBLEM*.

*On a given straight line to describe a segment of a circle, containing an angle equal to a given rectilineal angle*.

Let *AB* be the given straight line, and *C* the given rectilineal angle: it is required to describe, on the given straight line *AB*, a segment of a circle containing an angle equal to the angle *C*.

First, let the angle *G* be a right angle.

Bisect *AB* at *F*, [1. 10.

and from the centre *F*, at the distance *FB*, describe the semicircle *AHB*.

Then the angle *AHB* in a semicircle is equal to the right angle C. [III. 31.
But if the angle *C* be not a right angle, at the point *A*, in the straight line *AB*, make the angle *BAD* equal to the angle *C*; [I. 23.

from the point *A*, draw *AE* at right angles to *AD*;[I.ll.

bisect *AB* at *F*; [I. 10. from the point *F*, draw *FG* at right angles to *AB* [1. 11.

and join *GB*.

Then, because AF is equal to BF, [*Const*.

and FG is common to the two triangles *AFG*,*BFG*;

the two sides *AF*, *FG' are equal to the two sides *BF*, *FG*, each to each;*
and the angle

*AFG' is equal to the angle*BFG

*; [I.*Definition

*10.*

therefore the base

*AG*is equal to the base

*BG*; [I. 4.

and therefore the circle described from the centre

*G*, at the distance

*GA*, will pass through the point

*B*.

Let this circle be described; and let it be

*AHB*.

The segment

*AHB*shall contain an angle equal to the given rectilineal angle

*G*.

Because from the point *A*, the extremity of the diameter *AE*, *AD* is drawn at right angles to AE, [*Construction*.

therefore *AD* touches the circle. [III. 16. *Corollary*.

And because *AB* is drawn from the point of contact *A*, the angle *DAB* is equal to the angle in the alternate segment *AHB*. [III. 32.

But the angle *DAB* is equal to the angle *O*. [*Constr*.

Therefore the angle in the segment *AHB* is equal to the angle *C*. [*Axiom* 1.

*on the given straight line AB, the segment AHB of a circle has been described, containing an angle equal to the given angle C*. q.e.f.

*PROBLEM*.

*From a given circle to cut off a segment containing an angle equal to a given rectilineal angle*.

Let *ABC* be the given circle, and *D* the given rectilineal angle: it is required to cut off from the circle *ABC* a segment containing an angle equal to the angle *D*.

Draw the straight line *BF* touching the circle *ABC* at the point *B*; [III. 17.

and at the point *B*,in the straight line *BF*, make the angle *FBC* equal to the angle *D*. [I. 23.

The segment *BAC* shall contain an angle equal to the angle *D*.

Because the straight line *EF* touches the circle *ABC*, and *BC* is drawn from the point of contact *B*, [*Constr*.

therefore the angle *FBC* is equal to the angle in the alternate segment *BAC* of the circle. [III. 32.

But the angle *FBC* is equal to the angle *D*. [*Construction*.

Therefore the angle in the segment *BAC* is equal to the angle *D*. [*Axiom* 1.

Wherefore, *from the given circle ABC, the segment BAC has been cut off, containing an angle equal to the given angle D*. q.e.f.

*If two straight lines cut one another within a circle, the rectangle contained hy the segments of one of them shall be equal to the rectangle contained by the segments of the other*.
Let the two straight lines *AC*, *BD* cut one another at the point *E*, within the circle *ABCD*: the rectangle contained by *AE*,*EC* shall be equal to the rectangle contained by *BE*, *ED*.

If *AC* and *BD* both pass through the centre, so that *E* is the centre, it is evident, since *EA*, *EB*, *EC*, *ED* are all equal, that the rectangle *AE*, *EC* is equal to the rectangle *BE*,*ED*.

But let one of them, *BD*, pass through the centre, and cut the other *AC*, which does not pass through the centre, at right angles, at the point *E*. Then, if *BD* be bisected at *F*, *F* is the centre of the circle *ABCD*; join *AF*.

Then, because the straight line *BD* which passes through the centre, cuts the straight line *AC*, which does not pass through the centre, at right angles at the point *E*, [*Hypothesis*.*AE* is equal to *EC*. [III. 3.

And because the straight line *BD* is divided into two equal parts at the point *F*, and into two unequal parts at the point *E*, the rectangle *BE*, *ED*, together with the square on *EF*, is equal to the square on *FB*, [II. 5.

that is, to the square on AF.

But the square on *AF* is equal to the squares on *AE*,*EF*.[l.11. Therefore the rectangle *BE*, *ED*, together with the square on *EF*, is equal to the squares on *AE*, *EF*. [*Axiom* 1.

Take away the common square on *EF*, then the remaining rectangle *BE*, *ED*, is equal to the remaining square on *AE*, that is, to the rectangle *AE*, *EC*.

Next, let *BD*, which passes through the centre, cut the other *AC*, which does not pass through the centre, at the point *E*, but not at right angles. Then, if *BD* be bisected at *F*, *F* is the centre of the circle ABCD; join *AF*, and from *F* draw *FG* perpendicular to *AC*. [I. 12.

Then *AG* is equal to *GC*; [III. 3.

therefore the rectangle *AE*, *EC*, together with the square on *EG*, is equal to the square on *AG*. [II. 5.

To each of these equals add the square on *GF*;

then the rectangle *AE*, *EC*, together with the squares on *EG*, *GF*, is equal to the squares on *AG*, *GF*. [*Axiom* 2.

But the squares on *EG*, *GF* are equal to the square on *EF*;

and the squares on *AG*, *GF* are equal to the square on *AF*. [I. 47.

Therefore the rectangle *AE*, *EC*, together with the square on *EF*, is equal to the square on *AF*,

that is, to the square on *FB*.

But the square on *FB* is equal to the rectangle *BE*, *ED*, together with the square on *EF*. [II. 5.

Therefore the rectangle *AE*, *EC*, together with the square on *EF*, is equal to the rectangle *BE*, *ED*, together with the square on *EF*.

Take away the common square on *EF*;

then the remaining rectangle *AE*, *EG* is equal to the remaining rectangle *BE*, *ED*. [*Axiom* 3.

Lastly, let neither of the straight lines *AG*, *BD* pass through the centre. Take the centre *F*, [III. 1.

and through *E*, the intersection of the straight lines *AG*, *BD*, draw the diameter *GEFH*.

Then, as has been shewn, the rectangle *GE*, *EH* is equal to the rectangle *AE*, *EG*, and also to the rectangle *BE*, *ED*; therefore the rectangle *AE*, *EC* is equal to the rectangle *BE*, *ED*. [*Axiom* 1.

*if two straight lines*&c. q.e.d.

*If from any point without a circle two straight lines be drawn, one of which cuts the circle, and the other touches it; the rectangle contained by the whole line which cuts the circle, and the part of it without the circle, shall be equal to the square on the line which touches it*.

Let *D* be any point without the circle *ABC*, and let *DCA*, *DB* be two straight lines drawn from it, of which *DCA* cuts the circle and *DB* touches it: the rectangle *AD*, *DC* shall be equal to the square on *DB*.

First, let *DCA* pass through the centre *E*, and join *EB*. Then *EBD* is a right angle. [III. 18.

And because the straight line *AC* is bisected at *E*, and produced to *D*, the rectangle *AD*, *DC* together with the square on *EC* is equal to the square on *ED*. [II. 6.

But *EC* is equal to *EB*;

therefore the rectangle *AD*, *DC* together with the square on *EB* is equal to the square on 'ED*.*
But the square on

*ED*is equal to the squares on

*EB*,

*BD*, because

*EBD*is a right angle. [I. 47.

Therefore the rectangle

*AD*,

*DC*, together with the square on

*EB*is equal to the squares on

*EB*,

*BD*.

Take away the common square on

*EB*;

then the remaining rectangle

*AD*,

*DC*is equal to the square on

*DB*. [

*Axiom*3.

Next let

*DCA*not pass through the centre of the circle

*ABC*; take the centre

*E*; [III. 1.

from

*E*draw

*EF*perpendicular to

*AC*; [I. 12.

and hoin EB,EC,ED.

Then, because the straight line

*EF*which passes through the centre, cuts the straight line

*AC*, which does not pass through the centre, at right angles, it also bisects it; [III. 3.

therefore

*AF*is equal to

*FC*.

And because the straight line

*AC*is bisected at

*F*, and produced to

*D*, the rectangle

*AD*,

*DC*, together with the square on

*FC*, is equal to the square on

*FD*. [II. 6.

To each of these equals add the square on

*FE*.

Therefore the rectangle

*AD*,

*DC*together with the squares on

*CF*,

*FE*, is equal to the squares on

*DF*,

*FE*. [

*Axiom*2.

But the squares on

*CF*,

*FE*are equal to the square on

*CE*, because

*CFE*is a right angle; [I. 47.

and the squares on

*DF*,

*FE*are equal to the square on

*DE*.

Therefore the rectangle

*AD*,

*DC*, together with the square on

*CE*, is equal to the square on

*DE*.

But

*CE*is equal to

*BE*;

therefore the rectangle

*AD*,

*DC*, together with the square on

*BE*, is equal to the square on

*DE*.

But the square on

*DE*is equal to the squares on

*DB*,

*BE*, because

*EBD*is a right angle. [I. 47.

Therefore the rectangle

*AD*,

*DC*, together with the square on

*BE*, is equal to the squares on

*DB*,

*BE*.

Take away the common square on

*BE*;

then the remaining rectangle

*AD*,

*DC*is equal to the square on

*DB*. [

*Axiom*3.

Wherefore, *if from any point* &c. q.e.d.

*AB*,

*AC*, the rectangles contained by the whole lines and the parts of them without the circles are equal to one another; namely, the rectangle

*BA*,

*AE*is equal to the rectangle

*CA*,

*AF*; for each of them is equal to the square on the straight line

*AD*, which touches the circle.

*THEOREM*.

*If from any point without a circle there he drawn two straight lines, one of which cuts the circle, and the other meets it, and if the rectangle contained in the whole line which cuts the circle, and the part of it without the circle, be equal to the square on the line which meets the circle, the line which meets the circle shall touch it*.

Let any point *D* be taken without the circle *ABC*, and from it let two straight lines *DCA*, *DB* be drawn, of which *DCA* cuts the circle, and *DB* meets it; and let the rectangle *AD*, *DC* bw equal to the square on *DB*: *DB* shall touch the circle.

Draw the straight line *DE*, touching the circle *ABC* [III. 17.

find *F* the centre, [III. 1.

and join *FB*, *FD*, *FE*.

Then the angle *FED* is a right angle. [III. 18.

And because *DE* touches the circle *ABC*, and *DCA* cuts it, the rectangle *AD*, *DC* is equal to the square on *DE*. [III. 36.

But the rectangle *AD*, *DC* is equal to the square on *DB*. [*Hyp*.

Therefore the square on *DE* is equal to the square on *DB* [*Ax*.1.

therefore the straight line *DE* is equal to the straight line *DB*.

And *EF* is equal to *BF*; [I. *Definition* 15.

therefore the two sides *DE*, *EF* are equal to the two sides *DB*, *BF* each to each;

and the base *DF* is common to the two triangles *DEF*, *DBF*;

therefore the angle *DEF* is equal to the angle *DBF*. [I. 8.

But *DEF* is a right angle; [*Construction*.

therefore also DBF is a right angle.

And *BF*, if produced, is a diameter; and the straight line which is drawn at right angles to a diameter from the extremity of it touches the circle; [III. 16. *Corollary*.

therefore *DB* touches the circle *ABC*.

Wherefore, *if from a point* &c. q.e.d.