equal to the angle CBD, because the fide AD is equal to the Book IV. fide AB; therefore CBD, or DBA is equal to BCD; and confequently the three angles BDA, DBA, BCD are equal to one ano-8 5. I. ther. and because the angle DBC is equal to the angle BCD, the

h

fide BD is equal to the fide DC. but BD was made equal to CA, h. 6. 1. therefore also CA is equal to CD, and the angle CDA equal & 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. and BCD is equal to each of the angles BDA, DBA; each therefore of the angles BDA, DBA is double of the angle DAB. wherefore an isofceles triangle ABD is described having each of the angles at the bafe double of the third angle. Which was to be done,

PROP. XI. PROB.

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

то

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

a

Describe a an Isosceles triangle FGH having each of the angles a. 10. 4 at G, H double of the angle at F; and in the circle ABCDE infcribe the triangle ACD equiangular to the triangle FGH, fo . 2. 4× that the angle CAD 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. Bifect the angles ACD, CDA by the straight

A

Because each of the angles ACD, CDA is double of CAD, and are bifected by the ftraight lines CE, DB, the five angles DAC, ACE, ECD, CDB, BDA are equal to one another. but equal angles stand upon equal circumferences; therefore the five circumferen- d. 26. 3 ces AB, BC, CD, DE, EA are equal to one another. and equal

Book IV. circumferences are fubtended by equal straight lines; therefore the five ftraight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is

e. 29. 3.

f. 27.3.

a. 11. 4.

C. 18. 3.

qual to the angle AED. 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 fhewn that it is equilateral. Wherefore in the given circle an equilateral and equiangular pentagon has been infcribed, Which was to be done.

PROP. XII. PROB.

O defcribe an equilateral and equiangular pentagon about a given circle,

T°

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 propofition, be in the points A, B, C, D, E, fo that the circumfe rences AB, BC, CD, DE, EA are equal; and thro' the points A, b. 17. 3. B, C, D, E draw GH, HK, KL, LM, MG touching the circle; take the center F, and join FB, FK, FC, FL, FD. and because the ftraight line KL touches the circle ABCDE in the point C, to which FC is drawn from the center F, FC is perpendicular to KL; therefore each of the angles at C is a right angle. for the fame reafon, the angles at the points B, D are right angles. and because FCK is a right angle, the fquare of FK is equal to the fquares of FC, CK. for the fame reafon the fquare 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; the remaining square of CKis therefore equal to the remaining square

4. 47. I.

e

8.

of BK, and the straight line CK equal to BK. and because FB is Book IV. equal to FC, and FK common to the triangles BFK, CFK, the two BF, FK are equal to the two CF, FK; and the bafe BK is equal to the base KC; therefore the angle BFK is equal to the angle e. 3. 1. KFC, and the angle BKF to FKC. wherefore the angle BFC is double of the angle KFC, and BKC double of FKC. for the fame 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

f

H

is equal to the angle CFD. and
BFC is double of the angle KFC,
and CFD double of CFL; there-
fore the angle KFC is equal to the
angle CFL; and the right angle
FCK is equal to the right angle
FCL. therefore in the two triangles B
FKC, FLC, there are two angles

A

G

f. 27. 3.

FC, which is adjacent to the equal angles in each, is common to both; therefore the other fides fhall be equal to the other fides, g. 26. I. and the third angle to the third angle. therefore the ftraight line KC is equal to CL, and the angle FKC to the angle FLC. and becaufe KC is equal to CL, KL is double of KC. in the fame manner, it may be fhewn that HK is double of BK. and because BK is equal to KC, as was demonftrated, and that KL is double of KC, and HK double of BK, HK fhall be equal to KL. in like manner it may be fhewn that GH, GM, ML are each of them equal to HK or KL. therefore the pentagon GHKLM is equilateral. It is alfo equiangular; for fince 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 demonftrated; the angle HKL is equal to 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 five angles GHK, HKL, KLM, LMG, MGH being equal to one another, the pentagon GHKLM is equiangular. and it is equilateral, as was demonftrated; and it is defcribed about the circle ABCDE. Which was to be done.

Book IV.

a. 9. I.

b. 4. I.

C. 12. I.

d. 26. 1.

PROP. XIII. PROB.

O infcribe a circle in a given equilateral and equi.

Tangular pentagon.

b

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, and from the point F in which they meet draw the straight lines FB, FA, FE. therefore fince BC is equal to CD, and CF common to the triangles BCF, DCF, the two fides BC, CF are equal to the two DC, CF; and the angle BCF is equal to the angle DCF; therefore the base BF is equal to the base FD, and the other angles to the other angles, to which the equal fides are oppofite; 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 alfo 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 thefame manner it may be demonftrated that the angles BAE, AED are bisected by the straight lines AF,FE. H from the point F draw FG, FH, FK, FL, FM perpendiculars to the straight lines AB, BC, CD, DE, EA. and because the angle HCF is

the B

G

A

M

CK D

L

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; therefore the other fides fhall be equal 4, each to each; wherefore the perpendicular FH is equal to the perpendicular FK. in the fame 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 defcribed from the center F, at the distance of one of these five, shall pass thro' the extremities of the other four, and touch the straight

fines AB, BC, CD, DE, EA, because the angles at the points G, Book IV. H, K, L, M are right angles; and that a ftraight line drawn from n the extremity of the diameter of a circle at right angles to it,

touches the circle. therefore each of the straight lines AB, BC, e. 16. 3. CD, DE, EA touches the circle; wherefore it is inscribed in the pentagon ABCDE. Which was to be done.

T°

PROP. XIV. PROB.

O defcribe a circle about a given equilateral and
equiangular pentagon.

Let ABCDE be the given equilateral and equiangular pentagon; it is required to defcribe a circle about it.

Bisect the angles BCD, CDE by the ftraight lines CF, FD, a. 9. x, and from the point F in which they meet draw the straight lines

FB, FA, FE to the points B, A, E. It
maybe demonstrated, in the same man-
ner as in the preceding propofition,
that the angles CBA, BAE, AED are
bifected by the straight lines FB, FA, B
FE. and because the angle BCD is e-
qual to the angle CDE, and that FCD
is the half of the angle BCD, and
CDF the half of CDE; the angle FCD
is equal to FDC; wherefore the fide

b

F

D

CF is equal to the fide FD. in like manner it may be demon- b. 6. 1. ftrated that FB, FA, FE are each of them equal to FC or FD. therefore the five ftraight lines FA, FB, FC, FD, FE are equal to one another; and the circle defcribed from the center F, at the distance of one of them, shall pass thro' the extremities of the other four, and be defcribed about the equilateral and equiangular pentagon ABCDE. Which was to be done.