يا

b 6. 1.

Book IV. and EBA the half of ABC; the angle EAB is equal to the angle EBA; and the fide EA b to the fide EB. In the fame manner, it may be demonftrated, that the straight lines EC, ED are each of them equal to EA or EB; therefore the four ftraight lines EA, EB, EC, ED are equal to one another; and the circle defcribed from the centre E, at the distance of one of them, muft pafs through the extremities of the other three, and be defcribed about the fquare ABCD. Which was to be done.

a II. 2.

To

PRO P. X. PRO B.

defcribe an ifofceles triangle, having each of the angles at the bafe double of the third angle.

Take any ftraight line AB, and divide a it in the point C, fo that the rectangle AB, BC may be equal to the fquare of CA; and from the centre A, at the distance AB, defcribe the circle b1.4 BDE, in which placeb the ftraight line BD equal to AC,

E

which is not greater than the diameter of the circle BDE; 5.4 join DA, DC, and about the triangle ADC defcribe the circle ACD; the triangle ABD is fuch as is required, that is, each of the angles ABD, ADB is double of the angle BAD. Because the rectangle AB.BC is equal to the fquare of AC, and AC equal to BD, the rectangle AB.BC is equal to the fquare of BD; and becaufe from the point B without the circle ACD two ftraight lines BCA, BD are drawn to the circumference, one of which cuts, and the other meets the circle, and the rectangle AB.BC contained by the whole of the qutting line, and the part of it without the circle, is equal to the fquare of BD which meets it; the ftraight

d37.3 line BD touches d the circle

B

ACD. And because BD touches the circle, and DC is

drawn

e 32. 3.

f. 32. 1,

drawn from the point of contact D, the angle BDC is equal e Book IV. to the angle DAC in the alternate fegment of the circle; to each of these add the angle CDA; therefore the whole angle BDA is equal to the two angles CDA, DAC; but the exterior angle BCD is equal to the angles CDA, DAC; therefore alfo BDA is equal to BCD; but BDA is equalg g 5.1. to CDB, because the side AD is equal to the fide AB; therefore CBD, or DBA is equal to BCD; and confequently the three angles BDA, DBA, BCD, are equal to one another. And because the angle DBC is equal to the angle BCD, the fide BD is equal h to the fide DC; but BD was made equal h6. 1. to CA; therefore alfo CA is equal to CD, and the angle CDA equal g to the angle DAC; therefore the angles CDA, DAC together, are double of the angle DAC: but BCD is equal to the angles CDA, DAC; therefore alfo BCD is double of DAC. But BCD is equal to each of the angles BDA, DBA, and therefore each of the angles BDA, DBA is double of the angle DAB; wherefore an ifofceles triangle ABD is defcribed, having each of the angles at the base double of the third angle. Which was to be done.

"COR. I. The angle BAD is the fifth part of two right angles. For fince each of the angles ABD and ADB is equal to twice the angle BAD, they are together equal to four times BAD, and therefore all the three angles ABD, ADB, BAD, taken together, are equal to five times the angle BAD. But the three angles ABD, ADB, BAD are equal to two right angles, therefore five times the angle BAD is equal to two right angles; or BAD is the fifth part of two right angles."

"COR. 2. Becaufe BAD is the fifth part of two, or the tenth part of four right angles, all the angles about the centre. A are together equal to ten times the angle BAD, and may therefore be divided into ten parts each equal to BAD. And as these ten equal angles at the centre, must stand on ten equal arches, therefore the arch BD is one tenth of the circumference; and the ftraight line BD, that is AC, is therefore equal to the fide of an equilateral decagon infcribed in the circle BDE."

Book IV.

PROP. XI. PROB.

To infcribe an equilateral and equiangular pentagon in a given circle.

a 10. 4.

Let ABCDE be the given circle, it is required to infcribe an equilateral and equiangular pentagon in the circle ABCDE.

Describe a an ifofceles triangle FGH, having each of the angles at G, H, double of the angle at F; and in the circle b 2.4. ABCDE infcribe b the triangle ACD equiangular to the triangle FGH, fo that the angle CAD be equal to the angle at F, and each of the angles ACD, CDA equal to the angle at Gor H; wherefore each of the angles ACD, CDA is double of the angle CAD. c9. 1. Bifect the angles ACD, CDA by the ftraight lines CE, DB; and join AB, BC, DE, EA. ABCDE is the pentagon required. G

F

B

Because the angles ACD, CDA are each of them double of CAD, and are bifected by the straight lines CE, DB, the five angles DAC, ACE, ECD, CDB, BDA are equal to one anod 26. 3. ther; but equal angles ftand upon equal d circumferences; therefore the five circumferences AB, BC, CD, DE, EA are equal to one another and equal circumferences are fube 29. 3. tended by equal e ftraight lines; therefore the five ftraight lines AB, BC, CD, DE,EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also equiangular; because the circumference AB is equal to the circumfe

rence

rence DE: if to each be added BCD, the whole ABCD is equal Book IV. to the whole EDCB: and the angle AED stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is equal f to the angle f 27. 3. AED: for the fame 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 shown that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been infcribed. Which was to be done."

Otherwise.

"Divide the radius of the given circle, fo that the rectangle contained by the whole and one of the parts may be equal to the fquare of the other a. Apply in the circle, on a 11. 2. each fide of a given point, a line equal to the greater of these parts; then beach of the arches cut off will be one-tenth of sb 2. Cor. the circumference, and therefore the arch made up of both 10. 2. will be one-fifth of the circumference; and if the ftraight line fubtending this arch be drawn, it will be the fide of an equilateral pentagon infcribed in the circle."

PROP. XII. PROB.

TO defcribe 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 laft propofition, be in the points A, B, C, D, E, so that the arches

C

C 12. I. the point F draw c FG, FH, FK, FL, FM perpendiculars to the ftraight lines AB, BC, CD, DE, EA: and because the angle HCF is equal to KCF, and the right angle FHC equal to the right angle FKC; in the triangles FHC, FKC there are two angles of one equal to two angles of the other, and the fide FC, which is oppofite to one of the equal angles in each, is common to both; there26. 1. fore, the other fides fhall be equal, each to each; wherefore the perpendicular FH is equal to the perpendicular FK: in the fame manner it may be demonftrated, that FL, FM, FG are each of them equal to FH or FK: therefore the five ftraight lines FG, FH, FK, FL, FM are equal to one ther wherefore the circle defcribed from the centre F, at the distance of one of thefe five, will pass through the extremities of the other four, and touch the ftraight lines AB, BC, CD, DE, EA, because that the angles at the points G, H, K, L, M are right angles, and that a ftraight line drawn from the extremity of the diameter of a circle at right angles to • Cor.16.3. it, touches e the circle: therefore each of the ftraight lines AB, BC, CD, DE, EA touches the circle; wherefore the circle is infcribed in the pentagon ABCDE. Which was to be done.

:

ano

PROP.