Book III,

a 17.3.

18. 3.

€ 28. I.

34. I.

8 Io. I.

h. 31. I.

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

B

F

A

E

D

Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D draw2 FG, GH, HK, KF touching the circle; and because FG touches the circle ABCD, and EA is drawn from the center E to the point of contact A, the angles at A are right angles; for the fame reason, the angles at the points B, C, D are right angles; and because the angle AEB is a right G angle, as likewife is EBG, GH is parallel to AC; for the fame reason, AC is parallel to FK, and in like manner GF, HK may each of them be demonftrated to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms; and GF is therefore equal to HK, and GH to FK; and because AC is equal to BD, and that AC is equal to each of the two GH, FK; and BD to each of the two GF, HK, GH, FK are each of them equal to GF or HK ; therefore the quadrilateral figure FGHK is equilateral. It is alfo rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB d is likewife a right angle: In the fame manner, it may be fhewn that the angles at H, K, F are right angles; therefore the quadrilateral figure FGHK is rectangular, and it was demonstrated to be equilateral; therefore it is a fquare; and it is defcribed about the circle ABCD: Which was to be done.

H

PROP. VIII. PROB.

TO infcribe a circle in a given square.

C

K

Let ABCD be the given fquare; it is required to inscribe a circle in ABCD.

Bifect each of the fides AB, AD, in the points F, E, and through E draw EH parallel to AB or DC, and through F draw

FK

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 © 34. £. that AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the fides opposite 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 ftraight lines GE, GF, GH, GK, F are equal to one another; and the circle defcribed from the center G, at the distance of one of them, shall pass thro' the extremities of the other three, and touch the ftraight lines AB, BC, CD,

B

K

H

DA; because the angles at the points E, F, H, K are right dd 29. 1. angles, and that the straight line which is drawn from the ex

tremity of a diameter, at right angles to it, touches the circle; e 16. 3. therefore each of the 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 fquare; it is required to defcribe 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 straight line AC: In the fame manner, it may be demonftrated that the angles ABC, B BCD, CDA are feverally bifected by the ftraight lines BD, AC; therefore,

a 8. I.

because 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 fide EB: In the fame manner, it may be b 6.1.

demon.

Book IV. demonftrated that the ftraight 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 described from the center E, at the distance of one of them, fhall pass through the extremities of the other three, and be described about the fquare ABCD: Which was to be done.

a II, 2.

b 1.4.

€ 5.4.

& 37.3.

e 32. 3.

f 32. I.

PROP. X. PROB.

'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, fo that the rectangle AB, BC be equal to the fquare of CA; and from the center 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.

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 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; the straight line BD touches & 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 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;

B

D

but

but BDA is equal 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, are 8 5. A. equal to one another; and becaufe the angle DBC is equal to the angle BCD, the fide BD is equal to the fide DC; but BD h 6, 1, was made equal to CA; therefore alfo 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 ifofceles triangle ABD is defcribed, having each of the angles at the bafe 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 tri- 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 ACD, CDA by the ftraight lines CE, DB; and join AB,

c

F

B

E

BC, DE, EA. ABCDE G

H

is the pentagon required.

C 9. T.

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, ECÓ, CDB, BDA are equal to one another; but equal angles ftand upon equal circumferences; therefore the d 26. 3. five circumferences AB, BC, CD, DE, EA are equal to one

d

another:

€ 29.3.1

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 equal f 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 equiangu lar; 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.

£27.3°

a 11.4.

`b 17. 3.

c 18.3.

¿:47. I.

PROP. XII. PROB.

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 equala; and thro' the points A, 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 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 fquare of FK is equal to the fquares of FC, CK: For the fame reason, the fquare of FK is equal to the fquares of FB, BK: Therefore the fquares of FC, CK are equal to the fquares of FB, BK, of which the fquare of FC is equal to the fquare of FB; the remaining square of CK is therefore equal to

the

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