Join AC, BD, cutting one another in E:

and because DA is equal to AB, and AC common to the triangles DAC, BAC, (1. def. 30.)

the two sides DA, AC are equal to the two BA, AC, each to each; and the base DC is equal to the base BC;

wherefore the angle DAC is equal to the angle BAC, (1. 8.)
and the angle DAB is bisected by the straight line AC:

in the same manner it may be demonstrated that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC: therefore, because the angle DAB is equal to the angle ABC, (1. def. 30.) and that the angle EAB is the half of DAB, and EBA the half of ABC; therefore the angle EAB is equal to the angle EBA; (ax. 7.) wherefore the side EA is equal to the side EB: (1.`6.)

in the same manner it may be demonstrated, that the straight lines EC, ED are each of them equal to EA or EB:

therefore 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 one of them, will pass through the extremities of the other three, and be described about the square ABCD. Q.E.F.

PROPOSITION X. 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 in the point C, (II. 11.) so that the rectangle AB, BC may be equal to the square of CA; and 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.

Then the triangle ABD shall be such as is required,

that is, each of the angles ABD, ADB shall be double of the angle BAD.

E

B D

Join DC, and about the triangle ADC describe the circle ACD. (IV. 5.) And because the rectangle AB, BC is equal to the square of AC, and that AC is equal to BD, (constr.)

the rectangle AB, BC is equal to the square of BD: (ax. 1.) and because from the point B, without the circle ACD, two straight lines BCA, BD are drawn to the circumference, one of which cuts, and the other meets the circle, 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 of BD which meets it;

therefore the straight line BD touches the circle ACD: (111. 37.)

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

the angle BDC is equal to the angle DAC in the alternate segment of the circle: (111. 32.)

to each of these add the angle CDA;

therefore the whole angle BDA is equal to the two angles CDA, DAC: (ax. 2.)

but the exterior angle BCD is equal to the angles CDA, DAC ; (1. 32.) therefore also BDA is equal to BCD: (ax. 1.)

but BDA is equal to the angle CBD, (1. 5.)
because the side AD is equal to the side AB;

therefore CBD, or DBA, is equal to BCD; (ax. 1.)
and consequently the three angles BDA, DBA, BCD are equal to
one another:

and because the angle DBC is equal to the angle BCD,
the side BD is equal to the side DC: (1. 6.)
but BD was made equal to CA;

therefore also CA is equal to CD, (ax. 1.)

and the angle CDA equal to the angle DAC; (1. 5.) therefore the angles CDA, DAC together, are double of the angle DAC: but BCD is equal to the angles CDA, DAC; (1. 32.) therefore also BCD is double of DAC:

and BCD was proved to be equal to each of the angles BDA, DBA; therefore each of the angles BDA, DBA is double of the angle DAB. Wherefore an isosceles triangle ABD is described, having each of the angles at the base double of the third angle. Q. E. F.

PROPOSITION XI. 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.)

and in the circle ABCDE inscribe the triangle ACD equiangular to the triangle FGH, (Iv. 2.)

so that the angle CAD may be equal to the angle at F,

and each of the angles ACD, CDA equal to the angle at G or H; wherefore each of the angles ACD, CDA is double of the angle CAD. Bisect the angles ACD, CDA by the straight lines CE, DB; (1.9.) and join AB, BC, DE, EA.

Then ABCDE shall be the pentagon required.

Because each of the angles ACD, CDA is double of CAD,
and that they are bisected by the straight lines CE, DB;

therefore the five angles DAC, ACE, ECD, CDB, BDA are equal to one another:

but equal angles stand upon equal circumferences; (III. 26.) therefore the five circumferences AB, BC, CD, DE, EA are equal to one another:

and equal circumferences are subtended by equal straight lines; (111. 29.) therefore the five straight lines AB, BC, CD, DE, EA are equal to

one another.

Wherefore the pentagon ABCDE is equilateral.

It is also equiangular:

for, because the circumference AB is equal to the circumference DE, if to each be added BCD,

the whole ABCD is equal to the whole EDCB: (ax. 2.) but the angle AED stands on the circumference ABCD; and the angle BAE on the circumference EDCB; therefore the angle BAE is equal to the angle AED: (111. 27.) for the same reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED:

therefore the pentagon ABCDE is equiangular;

and it has been shewn that it is equilateral:

wherefore, in the given circle, an equilateral and equiangular pentagon has been described. Q.E. F.

PROPOSITION XII. 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 in the points A, B, C, D, E,

so that the circumferences AB, BC, CD, DE, EA are equal; (iv. 11.) 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. And because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the centre F,

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 FCK is a right angle,

the square of FK is equal to the squares of FC, CK: (1. 47.)

for the same reason, the square of FK is equal to the squares of FB, BK:

therefore the squares of FC, CK are equal to the squares of FB, BK; (ax. 1.)

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

therefore the remaining square of CK is equal to the remaining square of BK, (ax. 3.)

and the straight line CK equal to BK:

and because FB is equal to FC, and FK common to the triangles BFK, CFK,

the two BF, FK are equal to the two CF, FK, each to each;
and the base BK was proved equal to the base KC;
therefore the angle BFK is equal to the angle KFC, (1. 8.)
and the angle BKF to FKC: (1. 4.)

wherefore the angle BFC is double of the angle KFC,
and BKC double of FKC:

for the same reason, the angle CFD is double of the angle CFL, and CLD double of CLF:

and because the circumference BC is equal to the circumference CD, the angle BFC is equal to the angle CFD; (111. 27.) and BFC is double of the angle KFC,

and CFD double of CFL;

therefore the angle KFC 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 FKC, FLC, 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 the other sides are equal to the other sides, and the third angle to the third angle: (1. 26.)

therefore the straight line KC is equal to CL, and the angle FKC to the angle FLC:

and because KC is equal to CL,

KL is double of KC.

In the same manner it may be shewn that HK is double of BK:
and because BK is equal to KC, as was demonstrated,
and that KL is double of KC, and HK double of BK,
therefore HK is equal to KL: (ax. 6.)

in like manner it may be shewn that GH, GM, ML are each of them equal to HK, or KL:

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 KLM double of FLC, as was before demonstrated; therefore the angle HKL is equal to KLM: (ax. 6.) and in like manner it may be shewn,

that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM:

therefore the five angles GHK, HKL, KLM, LMG, MGH, being equal to one another,

the pentagon GHKLM is equiangular:

and it is equilateral, as was demonstrated;

and it is described about the circle ABCDE. Q.E.F.

PROPOSITION XIII. 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, in which they meet, draw the straight lines FB, FA, FE:

therefore since BC is equal to CD, (hyp.) and CF common to the triangles BCF, DCF,

the two sides BC, CF are equal to the two 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 FD, (1.4.)
and the other angles to the other angles, to which the equal sides
are opposite;

therefore the angle CBF is equal to the angle CDF:
and because the angle CDE is double of CDF,
and that CDE is equal to CBA, and CDF to CBF;
CBA is also double of the angle CBF;

therefore the angle ABF is equal to the angle CBF; wherefore the angle ABC is bisected by the straight line BF: in the same manner it may be demonstrated,

that the angles BAE, AED, are bisected by the straight lines AF, FE. From the point F, draw FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA: (1. 12.)

and because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC;

therefore in the triangles FHC, FKC, there are two angles 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 the other sides are equal, each to each; (1. 26.)

wherefore the perpendicular FH is equal to the perpendicular FK: in the same manner it may be demonstrated, 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:

wherefore the circle described from the centre F, at the distance of one of these five, will pass through the extremities of the other four, and touch the straight lines AB, BC, CD, DE, EA,

because the angles at the points G, H, K, L, M are right angles, and that a straight line drawn from the extremity of the diameter of a circle at right angles to it, touches the circle: (111. 16.)