equal to the angle CBD, because the side AD is equal to the Book IV. fide AB; therefore CBD, or DBA is equal to BCD; and consequently the three angles BDA, DBA, BCD are equal to one ano-8. 5. 1. ther, and because the angle DBC is equal to the angle BCD, the fide BD is equal h to the side DC. but BD was made equal to CA, h. 6. I. therefore also CA is equal to CD, and the angle CDA equal 6 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 also 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 isosceles triangle ABD is described having each of the angles at the base double of the third angle. Which was to be done.

10 inscribe an equilateral and equiangular pentagon To

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 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 inscribe the triangle ACD equiangular to the triangle FGH, so D. 2. 4that the angle CAD be equal

А. to the angle at F, and each of the angles ACD, CDA equal

F to the angle at Gor H; where

BK

E
fore each of the angles ACD,
CDA is double of the angle
CAD. Bisect c the angles

C. 9. In ACD, CDA by the straight lines CE, DB, and join AB, BC, DE, EA. ABCDE is the G H pentagon required.

Because each of the angles ACD, CDA is double of CAD, and are bisected by the straight lines CE, DB, the five angles DAC, ACE, ECD, CDB, BD A are equal to one another. but equal angles stand upon equal d 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 subtended by equal • straight lines ; therefore pod the five straight lines AB, BC, CD, DE, EA are equal to one e. 29. 3.

another. Wherefore the pentagon ABCDE is equilateral. It is
also equiangular ; because the

А.
circumference AB is equal to
the circumference DE, if to
each be added BCD, the whole

F
B

E
ABCD is equal 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 e-G H
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 shewn that it is
equilateral. Wherefore in the given circle an equilateral and equi-
angular pentagon has been inscribed, which was to be done.

O describe an equilateral and equiangular pentagon

about a given circle,

.

a. 11. 4:

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

straight line KL touches the circle ABCDE in the point C, to c. 18. 3.

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 same rea

fon, the angles at the points B, D are right angles. and because FCK 4. 47. I. is a right angle, the square of FK is equal d to the squares of FC,

CK. 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, of which the square of FC is equal to the square of FB; theremaining square of CKis therefore equal to the remainingsquare

f. 27. 3.

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 base BK is equal to the base KC; therefore the angle BFK is equal to the angle e. 8. 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 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 f to the angle CFD. and

G BFC is double of the angle KFC,

A and CFD double of CFL ; therefore the angle KFC is equal to the

M

F FCK is equal to the right angle FCL. therefore in the two triangles B FKC, FLC, there are two angles of one equal to two angles of the other, each to each, and the fide FC, which is adjacent to the equal angles in each, is common to both; therefore the other sides shall be equals to the other fides, S. 26. Ja and the third angle to the third angle. 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, HK fhall be equal to KL. in like manner it may be thewn that GH, GM, ML are each of them equal to HK ör 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; 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 demonstrated; and it is described about the circle ABCDE. Which was to be done.

Book IV.

PROP. XIII.

PROB.

a. 9. I.

b. 4. I.

O inscribe a circle in a given equilateral and equi.

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

Bisect a 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 since 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; there-
fore the base BF is equal to the base FD, and the other angles
to the other angles, to which the equal fides are oppofite; there-
fore the angle CBF is equal to the angle CDF. and because the an.
gle 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 an-

А.
gle ABF is equal to the angle CBF;
wherefore the angle ABC is bisect-

G M
ed by the straight line BF. in the
same manner it may be demonstrat-
ed that the angles BAE, AED are
bifected 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, C K D
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 side FC, which is op-
posite to one of the equal angles in each, is common to both;
therefore the other fides shall be equal“, each to each; 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 de-
scribed 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

B

C. 12. 1.

d. 26. I.

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 straight line drawn from the extremity of the diameter of a circle at right angles to it, touches e the circle. therefore each of the straight lines AB, BC, c. 16. 3. CD, DE, EA touches the circle ; wherefore it is inscribed in the pentagon ABCDE. Which was to be done.

Tequiangular pentagon.

O describe a circle about a given equilateral and

pentagon.

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

Bifect - the angles BCD, CDE by the straight lines CF, FD, a. 9. 1, and from the point F in which they meet draw the straight lines FB, FA, FE to the points B, A, E. It may be demonstrated, in the same man А. ner as in the preceding proposition, that the angles CBA, BAE, AED are bisected by the straight lines FB, FA,B

E FE. and because the angle BCD is equal to the angle CDE, and that FCD is the half of the angle BCD, and CDF the half of CDE; the angle FCD

D is equal to FDC; wherefore the fide 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 straight lines FA, FB, FC, FD, FE are equal to one another; and the circle described from the center F, at the distance of one of them, shall pass thro' the extremities of the other four, and be described about the equilateral and equiangular pentagon ABCDE. Which was to be done.