A E

D

FK parallel to AD or BC; therefore each of the figures AK, Book IV. KB, AH, HD, AG, GC, BG, GD is a parallelogram, and their oppofite fides are equal; and because AD is equal to AB, and e 34. 1. that AE is the half of AD, and 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 demonftrated that GH, GK are each of them equal to FG or GE; therefore the four straight lines GE, GF, GH, GK, F are equal to one another; and the circle defcribed from the centre 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,

B

G

K

C

H

DA; because the angles at the points E, F, H, K are right & 29. 1. angles, and that the straight line which is drawn from the extremity of a diameter, at right angles to it, touches the circle; e 16. 3. therefore each of the straight lines AB, BC, CD, DA touches the circle, which therefore is infcribed in the fquare ABCD. Which was to be done,

To

PROP. IX. PROB.

O defcribe a circle about a given square.

Let ABCD be the given fquare; it is required to describe a circle about it.

E

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; wherefore the angle DAC is equal to the angle BAC, and the angle DAB is bifected by the ftraight line AC: In the fame manner, it may be demonftrated that the angles ABC, BCD, CDA are feverally bifected by the straight lines BD, AC; therefore, because the

a 8. I.

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 fide EB: In the fame manner, it may beb 6. £.

demonftrated

Book IV. demonftrated that the ftraight lines EC, ED are each of thêm equal to EA or EB; therefore the four straight lines EA, EB, EC, ED are equal to one another; and the circle defcribed from the centre E, at the diftance of one of them, fhall pass through the extremities of the other three, and be defcribed about the fquare ABCD. Which was to be done.

à II. 2.

b I. 4.

C 5. 4.

d 37. 3.

e 32. 3.

f 32. I.

PROP. X. PROB.

O defcribe an ifofceles triangle, having each of the

Tangles at the bafe double of the third angle.

Take any ftraight line AB, and divide it in the point C, fo that the rectangle AB, BC be equal to the square of CA; and from the centre A, at the diftance AB, defcribe the circle BDE, in which place the ftraight 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.

A

E

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 cir cumference, 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; the ftraight line BD touches d the circle ACD; and because BD touches the circle, and DC is drawn from the point of contact D, the angle BDC is equal to the angle DAC in the alternate fegment of the circle; to each of thefe add the angle CDA; therefore the whole angie BDA is equal to the two angles CDA, DAC; but the exterior angle BCD is equal to the angles CDA, DAC; therefore alfo BĎA is equal to BCD;

B

D

but

but BDA is equals to the angle CBD, because the fide AD Book IV. is equal to the fide AB; therefore CBD, or DBA is equal to BCD; and confequently the three angles BDA, DBA, BCD, 8 5. I. 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 h 6. 1. BD was made equal to CA; therefore alfo CA is equal to CD, and the angle CDA equals 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 ifofceles triangle ABD is defcribed, having each of the angles at the base double of the third angle. Which was to be done.

T

O 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 a 10. 4. angles at G, H, double of the angle at F; and in the circle ABCDE infcribe the triangle ACD equiangular to the tria b 2. 4 angle 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 G or H; wherefore each of the angles ACD, CDA is double of the angle CAD. Bifect the angles

B

E

C 9. I

D

ACD, CDA by the ftraight lines CE, DB; and join AB,

BC, DE, EA. ABCDE is G

H

the pentagon required.

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 ftand upon equal circumferences; therefore the d 16. 8. five circumferences AB, BC, CD, DE, EA are equal to one

another :

W

€ 29. 3.

Book IV. another: And equal circumferences are fubtended by equal 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 alfo equiangular; because the circumference AB is equal to the circumference DE: If to each be added BCD, the whole ABCD is equal to the whole EDCB: And the angle AED ftands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is equalf to the angle ALD: For the fame reason, each of the angles ABC, BCD, CDE is equal to the angle BAE, or AED: Therefore the pentagon ABĊDE is equiangu lar; 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.

f 27. 3.

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

Ta

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

Let the angles of a pentagon, infcribed in the circle, by the laft propofition, be in the points A, B, C, D, E, fo that the circumferences AB, BC, CD, DE, EA are equal"; and thro' 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: And because the straight line KL touches the circle ABCDE in the point C, to which FC is drawn from the centre F, FC is perpendicular to KL; therefore each of the angles at C is a right angle: For the same reason, 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 reason, the fquare of FK is equal to the squares of FB, BK: Therefore the squares of FC, CK are equal to the 1quares 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

the remaining fquare of BK, and the ftraight line CK equal to Book IV. BK: And becaufe FB is 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 bafe KC; therefore the angle BFK is equal to the angle KFC, and the angle BKF to e 8. 1. 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 becaufe the circumference BC is equal to the circumference CD, the angle BFC 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 H
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

B

A

K

G

E

M

f 27. 3

is adjacent to the equal angles in each, is common to both; therefore the other fides fhall be equal to the other fides, and g 26. 1. 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 because KC is equal to CL, KL is double of KC: In the fame manner, it may be shown 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 fhown 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 shown, that each of the angles KHG, HGM, GML is equal to the angle HKL or KLM: There fore 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.