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Book IV.

a 17. 3.

b 18. 3.

c 28. 1.

d 34. 1.

PROP. VII. PROB.

To describe a square about a given circle.

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

F

Draw two diameters AC, BD of the circle ABCD, at right angles to one another, and through the points A, B, C, D, draw a FG, GH, HK, KF touching the circle; and because FG touches the circle ABCD, and EA is drawn from the centre E to the point of contact A, the angles at A are right angles; for the same reason the angles at the points B, C, D are right angles; and because the angle AEB is a right angle, as like- G wise is EBG, GH is parallel to AC; for the same reason, AC is parallel to FK, and in like manner GF, HK may each of them be de- B monstrated to be parallel to BED; therefore the figures GK, GC, AK, FB, BK are parallelograms; and

GF is therefore equal to HK, and H
GH to FK; and because AC is

A

E

D

C

K

a 10. 1.

equal to BD, and also 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 also rectangular; for GBEA being a parallelogram, and AEB a right angle, AGB is likewise a right angle: In the same manner, it may be shown 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 square; and it is described about the circle ABCD. Which was to be done.

PROP. VIII. PROB.

To inscribe a circle in a given square.

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

Bisect each of the sides AB, AD, in the points F, E, and b 31. 1. through E draw EH parallel to AB or DC, and through F

draw FK parallel to AD or BC; therefore each of the figures Book IV. AK, KB, AH, HD, AG, GC, BG, GD, is a parallelogram, and their opposite sides are equal; and because AD is equal c 34. 1. to AB, and because AE is the half of AD, and AF the half of AB, AE is equal to AF; wherefore the sides opposite to these are equal, viz. FG to GE; in the same manner, it may be demonstrated, that GH, GK are each of them equal to FG or GE; therefore the four F straight lines GE, GF, GH, GK are equal to one another; and the circle described from the centre G, at the distance of one of them, will

A

E

D

G

K

pass through the extremities of the B H other three, and will also touch the

C

straight lines AB, BC, CD, DA; because the angles at the points E, F, H, K are right angles, and because the straight d 29. 1. line which is drawn from the extremity of a diameter, at right angles to it, touches the circle ; therefore each of the straight e Cor.16.3. lines AB, BC, CD, DA touches the circle, which therefore is inscribed in the square 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 describe a circle about it.

a

E

a 8. 1.

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 sides DA, AC are equal to the two BA, AC, and the base DC is equal to the base BC; wherefore the angle DAC is equal to the angle BAC, and the angle DAB is bisected by the straight line AC: In the same manner, it may be demonstrated, that the angles ABC, BCD, CDA are severally bisected by the straight lines BD, AC; therefore, because the angle DAB is equal to the angle ABC, and the angle EAB is the EBA the half of ABC; the angle EAB is EBA; wherefore the side EA is equal to

B

half of DAB, and
equal to the angle
the side EB: In b 6. 1.

Book IV. the same manner, it may be demonstrated, that the straight lines EC, ED are each of them equal to EA, or EB; therefore, the four straight lines EA, ÉB, EC, ED are equal to one another; and the circle described from the centre E, at the distance of one of them, must pass through the extremities of the other three, and be described about the square ABCD. Which was to be done.

a 11.2.

b l. 4.

c 5. 4.

d 37. 3.

e 32. 3.

f 32. I.

PROP. X. PROB.

To describe an isosceles triangle, having each of the angles at the base double the third angle.

b

Take any straight line AB, and divide it in the point C, so that the rectangle AB, BC, may be equal to the square of CA; and from the centre A, at the distance AB, describe 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 describe the circle ACD; the triangle ABD is such as is required, that is, each of the angles ABD, ADB is double the angle BAD.

A

E

Because the rectangle AB, BC is equal to the square of AC, and AC is equal to BD, the rectangle AB, BC is equal to the square of BD; and because from the point B without the circle ACD, two straight lines BCA, BD, are drawn to the circumference, one of which cuts, and the other meets the circle, and the rectangle AB, BC contained by the whole of the cutting line, and the part of it without the circle, is equal to the square 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 segment 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 to the angles CDA, DAC; therefore also BDA is

B D

equal to BCD; but BDA is equal to the angle CBD, be- Book IV. cause the side AD is equal to the side AB; therefore CBD, or DBA is equal to BCD; and consequently the three angles g 5. 1. BDA, DBA, BCD, are equal to one another; and because the angle DBC is equal to the angle BCD, the side BD is equal to the side 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 angles DAC; therefore the angles CDA, DAC together, are double the angle DAC: But BCD is equal to the angles CDA, DAČ; therefore also BCD is double DAC, and BCD is equal to each of the angles BDA, DBA: each therefore of the angles BDA, DBA is double the angle DAB; wherefore an isosceles triangle ABD is described, having each of the angles at the base double the third angle. Which was to be done.

PROP. XI. PROB.

To inscribe an equilateral and equiangular pentagon in a given circle.

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

b

A

F

B

E

a 10. 4.

Describe a an isosceles triangle FGH, having each of the angles at G, H, double the angle at F; and in the circle ABCDE inscribe the triangle ACD equiangular to the tri- b 2.4. angle FGH, so 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 the angles CAD. Bisect c the angles ACD, CDA by the straight lines CE, DB: and join AB, BC, DE, EA. ABCDE is the pentagon required.

G

H

c 9. 1.

D

Because the angles ACD, CDA are each of them double 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; d 26. 3.

d

e

Book IV. therefore the five circumferences AB, BC, CD, DE, EA are equal to one another: And equal circumferences are subtended e 29. 3. by equal straight lines; therefore the five straight lines AB, BC, CD, DE, EA are equal to one another. Wherefore the pentagon ABCDE is equilateral. It is also 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 stands on the circumference ABCD, and the angle BAE on the circumference EDCB; therefore the angle BAE is equal 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 shown that it is equilateral. Wherefore, in the given circle, an equilateral and equiangular pentagon has been inscribed. Which was to be done.

f 27. 3.

a 11. 4.

b 17. 3.

c 18. 3.

d 47. l.

PROP. XII. PROB.

To describe 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.

a

Let the angles of a pentagon, inscribed in the circle, by the last proposition, be in the points A, B, C, D, E, so that the circumferences AB, BC, CD, DE, EA are equal; and through 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 square of FK is equal 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; the remaining square of CK is therefore equal to

d

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