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

*BOOK IV*.

1. A rectilneal figure is said to be inscribed in another rectilineal figure, when all the angles of the inscribed figure are on the sides of the figure in which it is inscribed, each on each.

2. In like manner, a figure is said to be described about another figure, when all the sides of the circumscribed figure pass through the angular points of the figure about which it is described, each through each.

3. A rectilineal figure is said to be inscribed in a circle, when all the angles of the inscribed figure are on the circumference of the circle.

4. A rectilineal figure is said to be described about a circle, When each side of the circumscribed figure touches the circumference of the circle.

5. In like manner, a circle is said to be inscribed in a rectilineal figure, when the circumference of the circle ouches each side of the figure. 6. A circle is said to be described about a rectilineal figure, when the circircumference of the circle passes through all the angular points of the figure about which it is described.

7. A straight line is said to be placed in a circle, when the extremities of it are in the circumference of the circle.

*PROBLEM*.

*In a given circle, to place a straight line, equal to a given straight line, which is not greater than the diameter of the circle*.

Let *ABC* be the given circle, and *D* the given straight line, not greater than the diameter of the circle: it is required to place in the circle *ABC*, a straight line equal to *D*.

Draw *BC*, a diameter of the circle *ABC*.

Then, if *BC* is equal to *D*, the thing required is done; for in the circle *ABC*, a straight line is placed equal to *D*.

But, if it is not, *BC* is greater than *D*. [*Hypothesis*.

Make *CE* equal to *D*, and from the centre *C*, at the distance *CE*, describe the circle *AEF*, and join *CA*.

Then, because *C* is the centre of the circle *AEF*, *CA* is equal to *CE*; [I. *Definition* 15.

but *CE* is equal to *D*; [*Construction*.

therefore *CA* is equal to *D*. [*Axiom* 1.

*in the circle ABC, a straight line CA is placed equal to the given straight line D, which is not greater than the diameter of the circle*. q.e.f.

*PROBLEM*.

*In a given circle, to inscribe a triangle equiangular to a given triangle*.

Let *ABC* be the given circle, and *DEF* the given triangle: it is required to inscribe in the circle *ABC' a triangle equiangular to the triangle *DEF*.*

Draw the straight line *GAH* touching the circle at the point *A*; [III. 17. at the point *A*, in the straight line *AH*, make the angle *GAB* equal to the angle *HAC*;[I 23.

and, at the point *A*, in the straight line *AG*, make the angle *GAB* equal to the angle *DFE*; and join *BC*. *ABC* shall be the triangle required.

Because *GAH* touches the circle *ABC*, and *AC* is drawn from the point of contact *A*, [*Construction*.

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

But the angle *HAC* is equal to the angle *DEF*. [*Constr*.

Therefore the angle *ABC* is equal to the angle *DEF*. [*Ax*.l.

For the same reason the angle *ACB* is equal to the angle *DFE*.

Therefore the remaining angle *BAC* is equal to the remaining angle *EDF*. [I. 32, *Axioms* 11 and 3.

Wherefore *the triangle ABC is equiangular to the triangle DEF, and it is inscribed in the circle ABC*. q.e.f.

*PROBLEM*.

*About a given circle, to describe a triangle equiangular to a given triangle*. Let *ABC* be the given circle, and *DEF* the given triangle: it is required to describe a triangle about the circle *ABC*, equiangular to the triangle *DEF*.

Produce *EF* both ways to the points *G*, *H*; take *K* the centre of the circle *ABC*; [III. 1.

from *K* draw any radius *KB*;

at the point *K*, in the straight line *KB*, make the angle *BKA* equal to the angle *DEG*, and the angle *BKC* equal to the angle *DFH*; [I. 23.

and through the points *A*, *B*, *C*, draw the straight lines *LAM*, *MBN*, *NCL*, touching the circle *ABC*. [III. 17.*LMN* shall be the triangle required.

Because *LM*, *MN*, *NL* touch the circle *ABC* at the points *A*, *B*, *C*, [*Construction*.

to which from the centre are drawn *KA*, *KB*, *KC*,

therefore the angles at the points *A*,*B*,*C* are right angles.[III.18.

And because the four angles of the quadrilateral figure *AMBK* are together equal to four right angles,/br>
for it can be divided into two triangles,

and that two of them *KAM*, *KBM* are right angles,

therefore the other two *AKB*, *AMB* are together equal to two right angles. [*Axiom*. 3.

But the angles *DEG*,*DEF* are together equal to two right angles. [I. 13.

Therefore the angles *AKB*, *AMB* are equal to the angles *DEG*, *DEF*;

of which the angle *AKB* is equal to the angle *DEG*; [*Constr*.

therefore the remaining angle *AMB* is equal to the remaining angle *DEF*. [*Axiom* 3.

In the same manner the angle *LNM* may be shewn to be equal to the angle *DFE*.

Therefore the remaining angle *MLN* is equal to the remaining angle *EDF*. [I. 32, Axioms 11 and 3. Wherefore *the triangle LMN is equiangular to the triangle DEF, and it is described about the circle ABC*. q.e.f.

*PROBLEM*.

*To inscribe a circle in a given triangle*.

Let *ABC* be the given triangle: it is required to inscribe a circle in the triangle *ABC*

Bisect the angles *ABC*, *ACB*, by the straight lines *BD*, *CD* meeting one another at the point *D*; [I.9

and from *D* draw *DE*, *DF*, *DG* perpendiculars to *AB*, *BC*, *CA*. [1.12.

Then, because the angle *EBD* IS equal to the angle *FBD*, for the angle *ABC* is bisected by *BD* [*Construction*.

and that the right angle *BED* is equal to the right angle *BFD*; [*Axiom* 11.

therefore the two triangles *EBD*, *FBD* have two angles and the side *BD*, which is opposite to one of the equal angles in each, is common to both;

therefore *DE* is equal to *DF*.

For the same reason *DG* is equal to *DF*.

Therefore *DE* is equal to *DG*. [*Axiom* 1.

Therefore the three straight lines *DE*, *DF*, *DG* are equal to one another, and the circle described from the centre *D*, at the distance of any one of them, will pass through the extremities of the other two;

and it will touch the straight lines *AB*, *BC*, *CA* because the angles at the points *E*, *F*, *G* are right angles, and the straight line which is drawn from the extremity of a dia- meter at right angles to it, touches the circle. [III. 16. *Cor*.

Therefore the straight line *AB*,*BC*,*CA* do each of them touch the circle, and therefor the circle is inscribed in the triangle *ABC*.

*a circle has been inscribed in the given triangle*. q.e.f.

*PROBLEM*.

*To describe a circle about a given triangle*.

Let *ABC* be the given triangle: it is required to describe a circle about *ABC*.

Bisect *AB*, *AC* at the points *D*, *E*; [I. 10.

from these points draw *DF*, *EF*, at right angles to *AB*,*AC*; [I. 11.*DF*, *EF*, produced, will meet one another;

for if they do not meet they are parallel,

therefore *AB*, *AC*, which are at right angles to them are parallel; which is absurd:

let them meet at *F*, and join *FA*; also if the point *F* be not in *BC*, join *BF*, *CF*.

Then, because *AD* is equal to *BD* [*Construction*.

and *DF* is common, and at right angles to *AB*, therefore the base *FA* is equal to the base *FB*. [I. 4.

In the same manner it may be shewn that *FC* is equal to *FA*.

Therefore *FB* is equal to *FC*; [*Axiom 1*.

and *FA*, *FB*, *FC* are equal to one another.

Therefore the circle described from the centre *F*, at the distance of any one of them, will pass through the extremities of the other two, and will be described about the triangle *ABC*.

Wherefore *a circle has been described about the given triangle*. q.e.f.

Corollary. And it is manifest, that when the centre
of the circle falls within the triangle, each of its angles is less than a right angle, each of them being in a segment greater than a semicircle; and when the centre is in one of the sides of the triangle, the angle opposite to this side, being in a semicircle, is a right angle; and when the centre falls without the triangle, the angle opposite to the side beyond which it is, being in a segment less than a semicircle, is greater than a right angle. [III. 31.

Therefore, conversely, if the given triangle be acuteangled, the centre of the circle falls within it; if it be a right-angled triangle, the centre is in the side opposite to the right angle; and if it be an obtuse-angled triangle, the centre falls without the triangle, beyond the side opposite to the obtuse angle.

*PROBLEM*.

*To inscribe a square in a given circle*.

Let *ABCD* be the given circle: it is required to inscribe a square in *ABCD*.

Draw two diameters *AC*, *BD* of the circle *ABCD*, at right angles to one another; [III. 1, I. 11.

and join *AB*, *BC*, *CD*, *DA*. The figure *ABCD* shall be the square required.

Because *BE* is equal to *DE*, for *E* is the centre;

and that *EA* is common, and at right angles to *BD*;

therefore the base *BA* is equal to the base 'DA*. [I. 4.*
And for the same reason

*BC*,

*DC*are each of them equal to

*BA*, or

*DA*.

Therefore the quadrilateral figure

*ABCD*is equilateral.

It is also rectangular.

For the straight line *BD* being a diameter of the circle *ABCD*, *BAD* is a semicircle; [*Construction*.

therefore the angle *BAD* is a right angle. [III. 31.

For the same reason each of the angles *ABC*, *BCD*, *CDA* is a right angle;

therefore the quadrilateral figure *ABCD* is rectangular.

And it has been shewn to be equilateral; therefore it is a square.

Wherefore*a square has been inscribed in the given circle*, q.e.f.

*PROBLEM*.

*To describe a square about a given circle*.

Let *ABCD* be the given circle: it is required to describe a square about it.

Draw two diameters *AC*, *BD* of the circle *ABCD*, at right angles to one another; [III. 1, 1. 11.

and through the points *A*,*B*,*C*,*D*, draw *FG*, *GH*, *HK*, *KF* touching the circle. [III. 17.

The figure *GHKF* shall be the square required.

Because *FG* touches the circle *ABCD*, and *EA* is drawn from the centre *E* to the point of contact *A*, [*Construction*.

therefore the angles at *A* are right angles. [III. 18.

For the same reason the angles at the points *B*, *C*, *D* are right angles.

And because the angle *AEB* is a right angle, [Construction.

and also the angle *EBG* is a right angle, therefore *GH* is parallel to *AC*. [I. 28.

For the same reason *AC* is parallel to FK*.*

In the same manner it may be shewn that each of the lines *GF*, *HK* is parallel to *BD*.

Therefore the figures *GK*, *GC*, *CF*, *FB*, *BK* are parallelograms;

and therefore *GF* is equal to *HK*, and *GH* to *FK*. [I. 34.

And because *AC* is equal to *BD*,

and that *AC* is, equal to each of the two *GH*, *FK*,

and that *BD* is equal to each of the two *GF*, *HK*,

therefore *GH*, *FK* are each of them equal to *GF*, or *HK*;

therefore the quadrilateral figure *FGHK* is equilateral.

It is also rectangular.

For since *AEBG* is a parallelogram, and *AEB* a right angle,

therefore *AGB* is also a right angle. [I. 34.

In the same manner it may be shewn that the angles at *H*, *K*, *F* are right angles;

therefore the quadrilateral figure *FGHK* is rectangular.

And it has been shewn to be equilateral; therefore it is a square.

Wherefore *a square has been described about the given circle*. q.e.f.

*PROBLEM*.

*To inscribe a circle in a given square*.

Let *ABCD* be the given square: it is required to inscribe a circle in *ABCD*.

Bisect each of the sides *AB*, *AD* at the points *F*, *E*; [I. 10.

through *E* draw *EH* parallel to *AB* or *DC*, and through *F* draw *FK* parallel to *AD* or *BC*. [I. 31.

Then each of the figures *AK*, *KB*, *AH*, *HD*, *AG*, *GC*, *BG*, *GD* is a right-angled parallelogram;

and their opposite sides are equal. [I. 34.

And because *AD* is equal to *AB*, [I. *Definition* 30.

and that *AE* is half of *AD*, and *AF* half of *AB*, [*Constr*. therefore *AE* is, equal to *AF*. [*Axiom* 7.

Therefore the sides opposite to these are equal, namely, *FG* equal to *GE*. [I. 34.

In the same manner it may be shewn that the straight lines *GH*, *GK* are each of them equal to *FG* or *GE*.

Therefore the four straight lines *GE*, *GF*, *GH*, *GK* are equal to one another, and the circle described from the centre *G*, at the distance of any one of them, will pass through the extremities of the other three;

and it will touch the straight lines *AB*, *BC*, *CD*, *DA*, because the angles at the points *E*, *F*, *H*, *K* are right angles, and the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle. [III. 16. Corollary.

Therefore the straight lines *AB*, *BC*, *CD*, *DA* do each of them touch the circle.

*a circle has been inscribed in the given square*. q.e.f.

*PROBLEM*.

*To describe a circle about a given square*.

Let *ABCD* be the given square: it is required to describe a circle about *ABCD*.

Join *AC*, *BD*, cutting one another at *E*.
Then, because *AB* is equal to *AD*,

and *AC* is common to the two triangles *BAC*, *DAC*;

the two sides *BA*, *AC* are equal to the two sides *DA*,*AC* each to each;

and the base *BC* is equal to the base *DC*;

therefore the angle *BAC* is equal to the angle *DAC*, [I. 8.

and the angle *BAD* is bisected by the straight line *AC*.

In the same manner it may be shewn that the angles *ABC*, *BCD',' *CDA* are severally bisected by the straight lines *BD*, *AC*.*

Then, because the angle *DAB* is equal to the angle *ABC*,

and that the angle *EAB* is half the angle *DAB*,

and the angle *EBA* is half the angle *ABC*,

therefore the angle *EAB* is equal to the angle *EBA*; [*Ax*. 7.

and therefore the side *EA* is equal to the side *EB*. [I. 6.

In the same manner it may be shewn that the straight lines *EC*, *ED* are each of them equal to *EA* or *EB*.

Wherefore the four straight lines *EA*, *EB*, *EC*, *ED* are equal to one another, and the circle described from the centre *E*, at the distance of any one of them, will pass through the extremities of the other three, and will be described about the square *ABCD*.

Wherefore *a circle has been described about the given sqaure*. q.e.f.

*PROBLEM*.

*To describe an isosceles triangle, having each of the angles at the base double of the third angle*.
Take any straight line *AB* and divide it at the point *C*, so that the rectangle *AB*, *BC* may be equal to the square on *AC*; [II. 11. from the centre *A*, at the distance *AB* describe the circle *BDE*, in which place the straight line *BD* equal to *AC*, which is not greater than the diameter of the circle *BDE*; [IV. 1.

and join *DA*. The triangle *ABD* shall be such as is required; that is, each of the angles *ABD*, *ADB* shall be double of the third angle *BAD*.

Join *DC*; and about the triangle *ACD* describe the circle *ACD*. [IV. 5.

Then, because the rectangle *AB*, *BC* is equal to the square on *AC*, [*Construction*.

and that *AC* is equal to *BD*, [*Construction*.

therefore the rectangle *AB*, *BC* is equal to the square on *BD*.

And, because from the point *B*, without the circle *ACD*, two straight lines *BCA*, *BD* are drawn to the circumference, one of which cuts the circle, and the other meets it,

and that the rectangle *AB*, *BC*, contained by the whole of the cutting line, and the part of it without the circle, is equal to the square on *BD* which meets it;

therefore the straight line *BD* touches the circle. [I II. 37.

And, because *BD* touches the circle *ACD*, and *DC* is drawn from the point of contact *D*,

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

To each of these add the angle *CDA*;

therefore the whole angle *BDA* is equal to the two angles *CDA*, *DAC*. [*Axiom* 2.

But the exterior angle *BCD* is equal to the angles *CDA* *DAC*. [1.32.

Therefore the angle *BDA* is equal to the angle *BCD* [[*Ax* .l. But the angle *BDA* is equal to the angle *DBA*, [I. 5.

because *AD* is equal to *AB*.

Therefore each of the angles *BDA*, *DBA*, is equal to the angle *BCD*. [Axiom 6.

And, because the angle *DBC* is equal to the angle *BCD*, the side *DB* is equal to the side *DC*; [1.6.

but *DB* was made equal to *CA*;

therefore *CA* is equal to *CD*, [*Axiom* 6.

and therefore the angle CAD is equal to the angle CDA. [I.5

Therefore the angles *CAD*, *CDA* are together double the angle *CAD*.

But the angle *BCD* is equal to the angles *CAD*, *CDA*. [I.32

Therefore the angle *BCD* is double of the angle *CAD*,

And the angle *BCD* has been shewn to be equal to each of the angles *BDA*, *DBA*;

therefore each of the angles *BDA*, *DBA* is double of the angle *BAD*.

Wherefore an isosceles triangle has been described, having each of the angles at the base double of the third angle, q.e.f.

*PROBLEM*.

*To inscribe an equilateral and equiangular pentagon in a given circle*.

Let *ABCDE* be the given circle: it is required to inscribe an equilateral and equiangular pentagon in the circle *ABCDE*.

Describe an isosceles triangle, *FGH*, having each of the angles at *G*, *H*, double of the angle at F [IV. 10.

in the circle *ABCDE*, inscribe the triangle *ACD*, equiangular to the triangle *FGH*, so that the angle *CAD* may
be equal to the angle at *F*, and each of the angles *ACD*, equal to the angle at *G* or *H* [IV. 2.

and therefore each of angles *ACD*,*ADC* is double of the angle *CAD*; bisect the angles *ACD*, *ADC* by the straight line *CE*,*DB*; [1.9.

and join AB, BC, AE, ED.*ABCDE* shall be the pentagon required.

For because each of the angles *ACD*, *ADC* is double of the angle *CAD*,

and that they are bisected by the straight lines *CE*, *DB*,

therefore the five angles *ADB*, *BDC*, *CAD*, *DCE*, *ECA*

equal to one another.

equal angles stand on equal arcs; [III. 26.

therefore the five arcs *AB*, *BC*, *CD*, *DE*, *EA* are equal to one another.

And equal arcs are subtended by equal straight lines; [III. 29.

therefore the five straight lines *AB*, *BC*, *CD*, *DE*, *EA* are equal to one another;

and therefore the pentagon *ABCDE* is equilateral.

It is also equiangular.

For, the arc *AB* is equal to the arc *DE*;

each of these add the arc *BCD*;

therefore the whole arc *ABCD* is equal to the whole arc *BCDE*. [*Axiom* 2.

And the angle *AED* stands on the arc *ABCD*, and the angle *BAE* on the arc *BCDE*. Therefore the angle *AED* is equal to the angle *BAE*. [III. 27.

For the same reason each of the angles *ABC*, *BCD*, *DE* is equal to the angle *AED* or *BAE*;

Therefore the pentagon ABCDE is equiangular.

And it has been shewn to be equilateral.

*an equilateral and equiangular pentagon has inscribed in the given circle*, q.e.f.

*PROBLEM*.

*To describe an equilateral and equiangular pentagon about a given circle*.

Let *ABCDE* be the given circle: it is required to describe an equilateral and equiangular pentagon about the circle *ABCDE*.

Let the angles of a pentagon, inscribed in the circle, by the last proposition, be at the points *A*, *B*, *C*, *D*, *E*, so that the arcs *AB*, *BC*, *CD*, *DE*, *EA* are equal; and through the points *A*, *B*, *C*, *D*, *E*, draw *GH*, *HK*, *KL*, *LM*, *MG* touching the circle. [III 17.

The figure *GHKLM* shall be the pentagon required.

Take the centre *F*, and join *FB*, *FK*, *FC*, *FL*, *FD*.

Then, because the straight line *KL* touches the circle *ABCDE* at the point *C* to which *FC* is drawn from the centre, therefore *FC* is perpendicular to *KL* [III. 18.

therefore each of the angles at *C* is a right angle.

For the same reason the angles at the points *B*, *D* are right angles.

And because the angle *FCK* is, a right angle, the square on *FK* equal to the squares on *FC*, *CK*. [I. 47.

For the same reason the square on *FK* is equal to the squares on *FB*, *BK*.

Therefore the squares on *FC*, *CK* are equal to the squares on *FB*,*BK*; [*Axiom* 1.

of which the square on *FC* is equal to the square on *FB*;

therefore the remaining square on *CK* is equal to the remaining square on *BK*, [*Axiom* 3.

and therefore the straight line *CK* is equal to the straight line *BK*. And because *FB* is equal to *FC*,

and *FK* is common to the two triangles *BFK*, *CFK* ;

the two sides *BF*, *FK* are equal to the two sides *CF*, *FK*,
each to each ;

and the base *BK* was shewn equal to the base *CK*;

therefore the angle *BFK* is equal to the angle *CFK*, [I. 8.

and the angle *BKF* to the angle *CKF*. [I. 4.

Therefore the angle *BFC* is double of the angle *CFK*,
and the angle *BKC* is double of the angle *CKF*.

For the same reason the angle *CFD* is double of the
angle *CFL*, and the angle *CLD* is double of the angle *CLF*.

And because the arc *BC* is equal to the arc *CD*,
the angle *BFC* is equal to the angle *CFD* ; [III. 27.

and the angle *BFC* is double of the angle *CFK*, and the
angle *CFD* is double of the angle *CFL* ;

therefore the angle *CFK* is equal to the angle *CFL*. [*Ax*. 7.

And the right angle *FCK* is equal to the right angle *FCL*.

Therefore in the two triangles *FCK*, *FCL*, there are two
angles of the one equal to two angles of the other, each to
each ;

and the side *FC*, which is adjacent to the equal angles in
each, is common to both ;

therefore their other sides are equal, each to each, and the
third angle of the one equal to the third angle of the other ;

therefore the straight line CK is equal to the straight line
*CL*, and the angle *FKC* to the angle *FLC*. [I. 26.

And because *CK* is equal to *CL*, *LK* is double of *CK*.

In the same manner it may be shewn that *HK* is double of *BK*.

And because *BK* is equal to *CK*, as was shewn,
and that *HK* is double of *BK*, and *LK* double of *CK*,

therefore *HK* is equal to *LK*. [*Axiom* 6.

In the same manner it may be shewn that *GH*,
txM, ML are each of them equal to *HK* or *LK*;
therefore the pentagon *GHKLM* is equilateral.

It is also equiangular.

For, since the angle *FKC* is equal to the angle *FLC*,

and that the angle *HKL* is double of the angle *FKC*, and the angle *KLM* double of the angle *FLC*, as was shewn,

therefore the angle *HKL* is equal to the angle *KLM*. [*Axiom* 6.

In the same manner it may be shewn that each of the angles *KHG*, *HGM*, *GML* is equal to the angle *HKL* or *KLM*;

therefore the pentagon *GHKLM* is equiangular. And it has been shewn to be equilateral.

Wherefore *an equilateral and equiangular pentagon has been described about the given circle*. q.e.f.

*PROBLEM*.

*To inscribe a circle in a given equilateral and equiangular pentagon*.

Let *ABCDE* be the given equilateral and equiangular pentagon: it is required to inscribe a circle in the pentagon *ABCDE*.

Bisect the angles *BCD*, *CDE* by the straight lines *CF*,*DF*; [1.9. and from the point *F*, at which they meet, draw the straight lines *FB*, *FA*, *FE*.

Then, because *BC* is equal to *DC* [*Hypothesis*.

and *CF* is common to the two triangles *BCF*, *DCF*;

the two sides *BC*, *CF* are equal to the two sides *DC*, *CF*, each to each;

and the angle *BCF* is equal to the angle *DCF*; [*Constr*.

therefore the base *BF* is equal to the base *DF*, and the other angles to the other angles to which the equal sides
are opposite ; [I. 4.

therefore the angle *CBF* is equal to the angle *CDF*.

And because the angle *CDE* is double of the angle
*CDF*, and that the angle *CDE* is equal to the angle *CBA*,
and the angle *CDF* is equal to the angle *CBF*,

therefore the angle *CBA* is double of the angle *CBF*;

therefore the angle. *ABF* is equal to the angle *CBF*;

therefore the angle *ABC* is bisected by the straight line *BF*.

In the same manner it may be shewn that the angles
*BAE*, *AED* are bisected by the straight lines *AF*, *EF*.

From the point *F* draw *FG*, *FH*, *FK*, *FL*, *FM* perpen-
diculars to the straight lines *AB*, *BC*, *CD*, *DE*, *EA*. [1. 12.

Then, because the angle *FCH* is equal to the angle
*FCK*,

and the right angle *FHC* equal to the right angle *FKC*;

therefore in the two triangles *FHC*, *FKC*, there are two
triangles of the one equal to two angles of the other, each to
each;

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

therefore their other sides are equal, each to each, and
therefore the perpendicular *FH* is equal to the perpen-
dicular *FK*. [I. 26.

In the same manner it may be shewn that *FL*, *FM*, *FG*
are each of them equal to *FH* or *FK*.

Therefore the five straight lines *FG*, *FH*, *FK*, *FL*, *FM* are
equal to one another, and the circle described from the
centre *F*, at the distance of any one of them will pass
through the extremities of the other four ;
and it will touch the straight lines *AB*, *BC*, *CD*, *DE*, *EA*,
because the angles at the points *G*, *H,* *K*, *L*, *M* are right
angles, [*Construction*.

and the straight line drawn from the extremity of a dia- "
meter, at right angles to it, touches the circle ; [III. 16.

Therefore each of the straight lines *AB*, *BC*, *CD*, *DE*, *EA*
touches the circle.

Wherefore *a circle has been inscribed in the given *
equilateral and equiangular pentagon*. q.e.f. PROPOSITION 14. *PROBLEM

*To describe a circle about a given equilateral and equiangular pentagon*.

Let *ABCDE* be the given equilateral and equiangular pentagon: it is required to describe a circle about it.

Bisect the angles *BCD*, *CDE* by the straight lines *CF*, *DF*; [1.9. and from the point *F*, at which they meet, draw the straight lines *FB*, *FA*, *FE*.

Then it may be shewn, as in the preceding proposition, that the angles *CBA*, *BAE*, *AED* are bisected by the straight lines *BF*, *AF*, *EF*.

And, because the angle *BCD* is equal to the angle *CDE*, and that the angle *FCD* is half of the angle *BCD*,

and the angle *FDC* is half of the angle *CDE*,

therefore the angle *FCD* is equal to the angle *FDC*, [*Ax*. 7.

therefore the side *FC* is equal to the side *FD*. [I. 6.

In the same manner it may be shewn that *FB*, *FA*, *FE* are each of them equal to *FC* or *FD*;

therefore the five straight lines *FA*, *FB*, *FC*, *FD*, *FE* are equal to one another, and the circle described from the centre *F*, at the distance of any one of them, will pass through the extremities of the other four, and will bo described about the equilateral and equiangular pentagon *ABCDE*.

Wherefore *a circle has been described about the given equilateral and equiangular pentagon*. q.e.f.

*PROBLEM*.

*To inscribe an equilateral and equiangular hexagon in a given circle*

Let *ABCDEF* be the given circle: it is requirerd to inscribe an equilateral and equiangular hexagon in it.

Find the centre *G* of the circle *ABCDEF*,
and draw the diameter AGD;

from the centre *D*, at the distance *DG* dscribe the circle *EGCH*;

join *EG*, *CG*, and produce them to the points *B*,*F*; and join *AB*, *BC*, *CD*, *DB*, *BF*, *FA*.

The hexagon *ABCDBF* shall be equilateral and equiangular.

For, because *G* is the centre of the circle *ABCDBF*, *GE* is equal to *GD*;

and because *D* is the centre of the circle *EGCH*, *DB* is equal to *DG*;

therefore *GE* is equal to *DE*,[*Axiom]] 1.*
and the triangle

*EGD*is equilateral;

therefore the three angles *EGD*, *GDE*, *DEG* are equal to one another. [I. 5. *Corollary*.

But the three angles of a triangle are together equal to two right angles; [I. 32.

therefore the angle *EGD* is the third part of two right angles.

In the same manner it may be shewn, that the angle *DGC* is the third part of two right angles.

And because the straight line *GC* makes with the straight line *EB* the adjacent angles *EGC*, *CGB* together equal to two right angles, [I. 13.

therefore the remaining angle *CGB* is the third part of two right angles;

therefore the angles *EGD*, *DGC*, *CGB* are equal to one aother.

And to these are equal the vertical opposite angles *BGA*,AGF,FGB. [1.15.

therefore the six angles *EGD*, *DGC*, *CGB*, *BGA*, *AGF*, *FGE* are equal to one another. But equal angles stand on equal arcs; [III. 26.

therefore the six arcs *AB*, *BC*, *CD*, *DE*, *EF*, *FA* are equal to one another.

And equal arcs are subtended by equal straight lines; [III.29.

therefore the six straight lines are equal to one another, and the hexagon is equilateral.

It is also equiangular. For, the arc *AF* is equal to the arc *ED';*
to each of these add the arc

*ABCD*;

therefore the whole arc

*FABCD*is equal to the whole arc

*ABCDE*;

and the angle

*FED*stands on the arc

*FABCD*,

and the angle

*FED*stands on the arc

*ABCDE*;

therefore the angle

*FED*is equal to the angle

*AFE*. [III. 27.

In the same manner it may be shewn that the other angles of the hexagon *ABCDEF* are each of them equal to the angle *AFE* or *FED*;

therefore the hexagon is equiangular.

And it has been shewn to be equilateral; and it is inscribed in the circle *ABCDEF*.

Wherefore *an equilateral and equiangular hexagon has been inscribed in the given circle*, q.e.f.

Corollary. From this it is manifest that the side of the hexagon is equal to the straight line from the centre, that is, to the semidiameter of the circle.

Also, if through the points*A*,

*B*,

*C*,

*D*,

*E*,

*F*, there be drawn straight lines touching the circle, an equilateral and equiangular hexagon will be described about the circle, as may be shewn from what was said of the pentagon; and a circle may be inscribed in a given equilateral and equir angular hexagon, and circumscribed about it, by a method like that used for the pentagon.

*PROBLEM*.

*To inscribe an equilateral and equiangular quindecagon in a given circle*.

Let *ABCD* be the given circle: it is required to inscribe an equilateral and equiangular quindecagon in the circle *ABCD*.

Let *AC* be the side of an equilateral triangle inscribed in the circle; [IV. 2.

and let *AB* be the side of an equilateral and equiangular pentagon inscribed in the circle. [IV. 11.

Then, of such equal parts as the whole circumference *ABCDEF* contains fifteen, the arc *ABC*, which is the third part of the whole, contains five, and the arc *AB*, which is tht; fifth part of the whole, contains three; therefore their difference, the arc *BC*, contains two of the same parts.

Bisect the arc *BC* at *E*; [III. 30.

therefore each of the arcs *BE*, *EC* is the fifteenth part of the whole circumference *ABCDF*.

Therefore if the straight lines *BE*, *EC* be drawn, and straight lines equal to them be placed round in the whole circle, [IV. 1.

an equilateral and equiangular quindecagon will be inscribed in it. q.e.f.

And, in the same manner as was done for the pentagon, if through the points of division made by inscribing the quindecagon, straight lines be drawn touching the circle, an equilateral and equiangular quindecagon will be described about it; and also, as for the pentagon, a circle may be inscribed in a given equilateral and equiangular quindecagon, and circumscribed about it.