AD or BC. therefore each of the figures AK, KB, AH, HD, AG, Book IV. GC, BG, GD is a parallelogram, and their oppofite fides are equal.m

F

A

E

D

G

K

B

H

C

and because AD is equal to AB, and that AE is the half of AD, and c. 34. 1. AF the half of AB; AE is equal to AF. wherefore the fides oppofite to these are equal, viz. FG to GE. in the fame manner it may be demonstrated that GH, GK are each of them equal to FG or GE. therefore the four straight lines GE, GF, GH, GK are equal to one another; and the circle defcribed from the center G, at the distance of one of them fhall pass thro' the extremities of the other three, and touch the ftraight lines AB, BC, CD, DA; because the angles at the points' E, F, H, K are right angles, and that d. 29. §4 the straight line which is drawn from the extremity of a diameter,

at right angles to it, touches the circle. therefore each of the e. 16. 3. ftraight lines AB, BC, CD, DA touches the circle, which therefore is infcribed in the fquare ABCD. Which was to be done.

PROP. IX. PROB.

To describe a circle about a given square.

Let ABCD be the given square; it is required to defcribe a circle about it.

Join AC, BD cutting one another in E. and because DA is equal to AB, and AC common to the triangles DAC, BAC, the two fides DA, AC are equal to the two BA, AC;

and the bafe DC is equal to the base BC; A

wherefore the angle DAC is equal to the angle BAC, and the angle DAB is bifected by the straight line AC. in the fame manner it may be demonstrated that the angles ABC, BCD, CDA are feverally bifected by the straight lines BD, AC. therefore be

B

E

D

a. 8. I.

cause the angle DAB is equal to the angle ABC, and that the angle EAB is the half of DAB, and EBA the half of ABC; the angle EAB is equal to the angle EBA; wherefore the fide EA is equal to the b. 61. fide EB. in the fame manner it may be demonftrated that the straight

G 3

b

lincs

Book IV. 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 center E, at the distance of one of them, fhall pass thro' the extremities of the other three, and be described about the fquare ABCD. Which was to be done.

2. 11. 2.

b. 1. 4.

C. 5. 4.

PROP. X. PROB.

To an

O defcribe an ifofceles 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 be equal to the fquare of CA; and from the center A, at the distance AB defcribe the circle BDE, in which place the straight line BD equal to AC, which is not greater than the diameter of the circle BDE; 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 that AC is equal to BD, the rectangle AB, BC is equal to the fquare of BD. and because 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 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 fquare of BD which meets it; d. 37. 3. the ftraight line BD touches the

A

E

circle ACD. and because BD

touches the circle, and DC is

B

D

drawn from the point of contact

e

e. 32. 3. D, the angle BDC is equal to the angle DAC in the alternate feg

f. 32. 1. 5. 5. 1.

ment 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 f to the angles CDA, DAC; therefore alfo BDA is equal to BCD. but BDA is equal to the angle CBD,

CBD, because the fide AD is equal to the fide AB; therefore Book IV.
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 to the h. 6. 1.
fide DC. but BD was made equal to CA, therefore alfo CA is equal

to CD, and the angle CDA equal & to the angle DAC. therefore the g. 5. 1.
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.

То infcribe an equilateral and equiangular pentagon

in a given circle.

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

Defcribe an Ifofceles triangle FGH having each of the angles at a. 10. 4. G, H double of the angle at F; and in the circle ABCDE infcribebb. 2. 4. the triangle ACD equiangular to the triangle FGH, fo that the

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, BDA are equal to one another. but equal angles stand upon equal circumferences; therefore the five circumferences d. 26.3. AB, BC, CD, DE, EA are equal to one another. and equal cir

e

Book IV. cumferences are fubtended by equal straight lines; therefore the five ftraight lines AB, BC, CD, DE, EA are equal to one another. e. 29. 3. Wherefore the pentagon ABCDE is equilateral. It is also equiangular; because the circumfe

rence AB is equal to the cir-
cumference DE, if to each be
added BCD, the whole ABCD
is equal to the whole EDCB.
and the angle AED stands on
the circumference ABCD,
and the angle BAE on the
circumference EDCB; there-

f. 27. 3. fore the angle BAE is equal f

3. 11.4.

A

to the angle AED. for the fame reafon, 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.

O 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 infcribed in the circle, by the last propofition, be in the points A, B, C, D, E, fo that the circumferences AB, BC, CD, DE, EA are equal 2; 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 which e. 18. 3. 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 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. for the fame reason the fquare of FK is equal to the fquares of FB, BK. therefore the squares of FC, CK are equal to the squares of FB, BK, of which the fquare of FC is equal to the fquare of FB; the remaining fquare of CK is therefore equal to the remaining fquare of BK,

d. 47 I.

d

and

f. 27.3.

and the straight line CK equal to BK. and because FB is equal to Book IV. 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 bafe KC; therefore the angle BFK is equal to the angle KFC, and the c. 8. 1. 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 angie BFC is equal f to the angle CFD. and EFC is double of the angle KFC, and CFD double of CFL; therefore 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 FKC, FLC, there are two angles of one equal to two angles of the other, each to

each, and the fide FC, which is

H

B

G

A

E

F

K

L

adjacent to the equal angles in each, is common to both; therefore

the other fides fhall be equal to the other fides, and the third g. 26. 1. 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 fame manner, it may be shewn that HK is double of BK. and becaufe 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 described about the circle ABCDE. Which was to be done.

PROP.