PROP. VIII. PROB L.

To infcribe a circle in a given fquare.

Let the given fquare be A B C D. It is required to infcribe a circle in the given square A B C D.

Divide [by 10. 1.] the fides A B, AD into equal parts, in the points F, E; and thro' E draw [by 31. 1.] E H parallel to A B or C D, and thro' F draw F K parallel to A D

or B C.

B

E

H

D

K

C

Then will AK, KB, AH, HD, AG, GC, BG, GD, be each of them parallelograms; and [by 34. 1.] their oppofite fides are equal. And because A D is equal to A B, and A E is one half of A D, and AF one half of AB; AE will be equal to AF: and the fides oppofite to them are equal; therefore F G is equal to G E. In like manner we demonftrate that GH, GK are each equal to F G, GE. Therefore the four right lines GE, GF, GH, G K are equal to one another; and fo a circle defcribed with the centre G, and either of the diftances GE, GF, GH, GK, will also país thro' the remaining points, and will touch the right lines A B, BC, CD, DA, because the angles at E, F, H, K are right angles: For if the circle cuts the right lines A B, B C, C D, DA, a right line drawn from the end of the diameter of a circle at right angles, will fall within the circle, which [by 16. 3.] is abfurd. Therefore a circle defcribed with the centre G, and either of the distances G E, GF, GH, GK will not cut the right lines A B, BC, CD, DA. It will therefore touch them, and [by 5. def. 4.] will be infcribed in the fquare A B C D....

Therefore a circle is infcribed in a given fquare. Which was to be done.

To circumfcribe a circle about a given Square.

Let the given fquare be A B C D. It is required to circumfcribe a circle about the square A B C D.

A

B

E

D

C

For join AC, BD, cutting one another in the point E. Then because D A is equal to A B, and A C is common, the two fides DA, A C are equal to the two fides B A, A C, and the base D C equal to the bafe BC; [by 8. 3.] the angle DAC will be equal to the angle BAC: Therefore the angle D A B is bifected by the right line A e. After the fame manner we demonstrate that each of the angles ABC, BCD, CDA, is bifected by the right lines AC, DB. Therefore because the angle D A B is equal to the angle ABC, and the angle E AB is one half of the angle DA B, and the angle E B A one half of the angle A B C; the angle E A B will be equal to the angle E BA: Wherefore [by 6. 1.] the fide E A is equal to the fide E B. In like manner we demonftrate that the right lines E C, E D are each equal to each of the right lines E A, E B. Therefore the four right lines EA, E B, EC, E D are equal to one another. Wherefore a circle described with the centre E, and either of the diftances E A, E B, E C, E D will pass thro' the rest of the points, and will be cirumfcribed about the fquare A B C D. Let it be defcribed, as A B C D.

Therefore a circle is circumfcribed about a given square, Which was to be done.

PROP. X. PROBL.

To make an ifofceles triangle having one of the angles at the bafe double to the remaining angle.

Let A B be any given right line, and [by 11. 2.] divide it in the point c, fo that the rectangle under AB, BC be equal to the fquare of ca; and about the centre A with the distance A B [by 3. poft.] describe a circle B DE, and [by 1. 4.] apply the right line B D in the circle B D E, which is not greater than the diameter, equal to CA, and joining DA, DC circumfcribe the circle A CD [by 5. 4.] about the triangle A C D.

Then because the rectangle under A B, B C is equal to the fquare of Ac; and AC is equal to BD; the rectangle

N 2

B

D

A

E

rectangle under A B, BC will be equal to the fquare of BD; and because the point B is taken without the circle ACD, and two right lines BCA, BD drawn from the the circle AC D, point B fall upon one of which cuts the circle, and the other meets it, and the rectangle under AB, BC is equal to the fquare of BD: the right line B D [by 37. 3.] will touch the circle a CD ; and therefore becaufe B D is a tangent, and D C is drawn from the point of contact D; the angle B D C [by 32. 3.] will be equal to the angle D A C in the alternate fegment of the circle; and fince the angle B D C is equal to DAC; add CDA, which is common to both; then the whole angle B D A is equal to the two angles e DA, DA C. But the outward angle BCD [by 32. 1.] is equal to CD A, DAC; and therefore BDA is equal to B C D. But the angle B DA [by 5. 1.] is equal to the angle C B D, because the fide A D is equal to the fide AB and therefore DBA will be equal to BCD. Wherefore the three angles B D A, D B A, BCD are equal to one another: And because the angle D B C is equal to the angle B C D, the fide B D [by 6. 1.] will be equal to the fide D C. But B D is made equal to CA; therefore AC is equal to CD; and so the angle CDA [by 5. 1.] is equal to the angle D AC: Therefore the angles CDA, DA C, taken together, are double to the angle D A C: But [by 32. 1.] the angle BCD alfo is equal to the angles C D A, DAC: Therefore B C D is double to D A C. But B C D is equal to either of the angles BD A, D BA; and therefore either B D A or D B A is double to D A B.

Therefore there is defcribe an ifofceles triangle A D B, having either of the angles at the base B D, double to the remaining angle. Which was to be done.

PROP. XI. PROB L.

To infcribe an equilateral and equiangular pentagon in a given circle c.

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

ABCDE.

Let

Let [by 10.4.] F G H be an ifofceles triangle having either of the angles at G, H double to the remaining angle at F and [by 2. 4.] infcribe the triangle A CD in the circle A B C D E equiangular to the triangle F G H ; fo that the angle C A D is equal to the angle F, the angles ACD,

F

G H

B

CDA equal to either cf the angles at G, H. Then either of the angles A CD, E CDA is double to the angle CAD. Now [by 9. 1.] bifact each of the angles A CD, CDA by the right lines CE, DB, and draw A B, B C, DE, EA.

Then because either of the angles A CD, CDA is double to the angle c A D, and they are bifected by the right lines C E, DB: the five angles DAC, ACE, ECD, CD B, BDA are equal to one another. But [by 26. 3.] equal angles ftand upon equal parts of the circumference. Therefore the five parts AB, BC, CD, DE, EA of the circumference are equal to one another. But [by 29. 3.] equal parts of the circumference are joined by equal right lines. Therefore the five right lines A B, B C, C D, DE, E A are equal to one another: Therefore ABCDE is an equilateral pentagon: I fay it is alfo equiangular. For because the part A B of the circumference is equal to the part DE: add the common part B C D, then will the whole A B C D, be equal to the whole EDC B. But the angle AED ftands upon the part A B C D of the circumference, and the angle BAE upon the part E D C B. Therefore [by 27. 3.] the angle BAE is equal to the angle A E D. By the fame reason each of the angles A B C, G C D, C D E is equal to either of the angles B A E, A E D. Therefore the pentagon A B C D E is equiangular. And it has been proved to be equilateral too.

Therefore an equilateral and equiangular pentagon is inscribed in a given circle. Which was to be done.

The following elegant way of defcribing a regular pentagon in a given circle (to be found in lib. i. of the great conftruction of Ptolemy) is in reality much fhorter than Euclid's way. But as its demonftration depends upon the thirteenth book, Euclid himself, if he had known this method, N 3

could

L

K

D

could not have put it into the fourth book, the materials being not yet prepared for it. The method is thus. Draw two diameters A C, B D of the I circle at right angles [by 11. 1.] meeting at the centre F. Bifect [by 10. 1.] the femidiameter A F in E. Join E, B. And upon A c from E lay off E G equal to E B. And draw

C

the right line BG: which will be the fide of the pentagon required. Then from B apply в I, BH equal to BG, as alfo IK, KL, LH: And the regular pentagon BIKLH will be inscribed in the circle A B C D.

Note also that F G is the fide of a regular decagon or figure of ten equal fides and angles infcribed in that circle. The demonftration easily follows from the ninth and tenth propofitions of the thirteenth book of Euclid.

To sircumfcribe an equilateral and equiangular pentagon about a given circle.

Let the given circle be A B C D E. It is required to circumfcribe an equilateral and equiangular pentagon about the circle A B C D E.

Let the points of the angles of a pentagon infcribed [by II. 4.] in the circle be fuppofed to be A, B, C, D, E, so that the parts of the circumference AB, B C, CD, DE, E A be equal: And [by 17. 3.] thro' the points A, B, C, D, E draw the right lines GH, HK, KL, LM, MG to touch the circle: find its centre F: and join F B, FK, FC, FL, F D.

Then because the right line KL touches the circle ABCDE in the point c, and F C is drawn from the centre F to the point of contact c; the line F c [by 18. 3.] will be perpendicular to K L. And fo each of the angles at c will be a right angle. By the fame reafon the angles at the points B, D are right angles. And because the angle FCK is a right angle, the fquare of F K [by 47. 1.] is equal to the squares of F C, CK. And by the fame reason the fquare of F K is equal to the fquares of F B, B K: Therefore the fquares of F C, C K are equal to the fquares of FB, BK; but the fquare of FC is equal to the fquare of FB: Therefore the remaining fquare of CK wil be

equal