PROP. X. PROB.

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, so that the rectangle AB, BC may be equal to the square of AC (2. 11); from the centre A, at the distance AB, describe the circle BDE, in which place the straight line BD equal to AC (4. 1), which is not greater than the diameter of the circle BDE, and join DA: the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double of the third angle BAD.

B

E

Join CD, and about the triangle ACD describe a circle: Then, because the rectangle AB, BC is equal to the square of AC, and that AC is equal to BD, therefore the rectangle AB, BC is equal to the square of 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, 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 BD touches the circle ACD (3. 37): And because BD touches the circle ACD, and that DC is drawn from the point of contact D cutting the circle, the angle BDC is equal to the angle CAD in the alternate segment of the circle (3. 32): To each of these add the angle CDA; therefore the whole angle BDA is equal to the two angles CAD, CDA: But the exterior angle BCD is equal to the angles CAD, CDA; therefore also the angle BDA is equal to the angle BCD: But the angle BDA is equal to the angle DBA, because AD is equal

to AB; therefore each of the angles BDA, DBA, is equal to the angle BCD: Now, because the angle DBA, or DBC, is equal to the angle BCD, the side DB is equal to the side DC: But DB was made equal to CA, therefore also CA is equal to CD, and the angle CAD to the angle CDA; and therefore the angles CAD, CDA are together double of the angle CAD: But the angle BCD is equal to the angles CAD, CDA; therefore also the angle BCD is double of the angle CAD: And the angle BCD is 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 ABD has been described, having each of the angles at the base double of the third angle. Q. E. F.

PROP. XI. PROB.

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.

F

Describe an isosceles triangle FGH, having each of the angles at G, H, double of the angle at F (4. 10); and in the circle ABCDE, inscribe the triangle ACD, equiangular to the triangle FGH (4. 2), so that the angle CAD may

be equal to the angle at F, and each of the angles ACD, ADC equal to the angle at G or H, and therefore each double of the angle CAD; bisect the angles ACD, ADC by the straight lines CE, DB, and join AB, BC, AE, ED: ABCDE is the pentagon required.

For, because each of the angles ACD, ADC is double of CAD, and that they are bisected by the straight lines CE, DB, therefore the five angles ADB, BDC, CAD, DCE, ECA, are equal to one another: But equal angles stand upon equal circumferences (3. 26); therefore the five circumferences, AB, BC, CD, DE, EA, are equal to one another: And equal circumferences are subtended by equal straight lines (3. 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, because the circumference AB is equal to the circumference DE, add to each the circumference BCD; therefore the whole circumference ABCD is equal to the whole circumference BCDE: And the angle AED stands on the circumference ABCD, and the angle BAE on the circumference BCDE; therefore the angle AED is equal to the angle BAE (3. 27): For the like reason, each of the angles ABC, BCD, CDE is equal to the angle AED or BAE; therefore the pentagon ABCDE is equiangular; and it has been shewn to be equilateral : Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed.

PROP. XII. PROB.

Q. E. F.

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; and through the points A, B, C, D, E, draw GH, HK, KL, LM, MG, touching the circle; take the centre F, and join FB, FK, FC, FL, FD.

Then, because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the centre, FC is perpendicular to KL (3. 18), and therefore each of the angles at C is a right angle:

For

the like 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, and, for the like 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, of which the square of FC is equal to the square of FB, and therefore the remaining square of CK is equal to the remaining square of BK, and the straight line CK to the straight line 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 is equal to the base CK-therefore the angle BFK is equal to the angle CFK, and the angle BKF to the angle CKF, and therefore the angle BFC is double of the angle CFK, and the angle BKC of the angle CKF: For the like reason, the angle CFD is double of the angle CFL, and the angle CLD of the angle CLF : But, because the circumference BC is equal to the circumference CD, the angle BFC is equal to the angle CFD; and BFC is double of CFK, and CFD of CFL; H therefore also the angle CFK is equal to the angle CFL: And the right angle FCK is equal to the right angle FCL: Therefore, in the two triangles FCK, FCL, there are two angles of one equal to two angles of the

B

G

K CL

E

M

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 (1. 26), each to each, and the third angle of the one to the third angle of the other; that is, CK is equal to CL, and the angle FKC to the angle FLC: And because CK is equal to CL, KL is double of CK: And in like manner it may be shewn that HK is double of BK: And because BK is equal to CK, as was demonstrated, and that HK is double of BK, and KL of CK, therefore also HK is equal to KL: And, 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, because, as has been shewn, the angle FKC is equal to the angle FLC, and the angle HKL is double of the angle FKC, and the angle KLM of the angle FLC, therefore the angle HKL is equal to the angle KLM : 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 pentagon GHKLM is equiangular and it has been shewn to be equilateral: Wherefore about the given circle has been described an equilateral and equiangular pentagon. Q. E. F.

PROP. XIII. PROB.

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