PROP. XXX. THEOR.

To cut a given right line into mean and extreme

ratio °.

Let A B be the given right line. It is required to cut the fame into mean and extreme ratio.

A

E D

B

For [by 46. 1.] upon A B defcribe the fquare BC; and C [by 29. 6.] apply the parallelogram CD to A c equal to BC redundant by the parallelogram A D fimilar to в C. F But BC is a fquare: Therefore AD will also be a fquare. H And becaufe B C is equal to CD, take away CE, which is common: mainder BF is equal to the remainder A D. But it is alfo equiangular to it: Wherefore [by 14. 6.] the fides of BF, AD about the equal angles are reciprocally proportional: Therefore as FE is to ED, fe is AE to E B. But [by 34. 1.] F E is equal to A C, that is, to A B, and ED equal to A E. Wherefore as A B is to A E, fo is AE to E B. But AB is greater than AE: Therefore A E is greater than E B.

Then the re

Therefore the right line AB is cut into mean and extreme ratio in E, the greater fegment being A E. Which was to be done.

Otherwise.

Let the given right line be a B.

It is required to cut the fame into mean and extreme ratio.

For [by 11. 2.] divide A B fo in c, that the rectangle under AB, BC be equal to the fquare of AC.

A

C B

Then because the rectangle under AB, BC is equal to the fquare of Ac; [by 17. 6.] it will be as A B to A c, fo is AC to CB. Therefore [by def. 3. 6.] AB is cut into mean and extreme ratio. Which was to be done.

This problem in fact is only a repetition of the eleventh propofition of the fecond book in other words. The first manner of folution here being indeed more difficult than that in the second book, and the latter the very fame as that in the fecond book.

PROP.

PROP. XXXI. THEOR.

In right angled triangles the figure defcribed upon the fide oppofite to the right angle is equal to both the figures defcribed upon the fides containing the right angle, which are fimilar to it, and have the fame fituation.

Let the right angled triangle be ABC, having the right angle BAC I fay the figure defcribed upon BC is equal to those described upon BA, AC being fimilar to it, and having the fame fituation.

Draw the perpendicular A D.

Then because AD is drawn from the right angle A of the right angled triangle ABC perpendicular to the base BC: [by 8. 6.] the triangles A B D, A DC are both fimilar to one another, and to the whole triangle ABC: 1 herefore it will be as c B is to BA, fo is A B to BD: But when three right lines are proportional; as the firft is to the third, fo [by 2. cor. C25. 6.] is the figure defcribed upon the firit to a figure finiilar and alike fituate, deTherefore as CB is to BD, fo

B

D

fcribed upon the fecond: is the figure described upon CB to a fimilar and alike fituate figure described upon B A. By the fame reason, as BC is to CD, fo is the figure defcribed upon BC, to the figure described upon c A: Therefore as Bc is to BD and DC together, fo is the figure defcribed upon BC to both the figures described upon BA, AC being fimilar and alike fituate to the figure defcribed upon BC. But BC is equal to BD, DC together: Therefore the figure defcribed upon BC is equal to both the figures together which are described upon BA, AC fimilar and alike fituate to the figure defcribed upon BC.

Therefore in right angled triangles, the figure described upon the fide oppofite to the right angle, is equal to both the figures together described upon the fides containing the right angle, which are fimilar to it, and have the fame fituation. Which was to be demonftrated,

Otherwife.

Otherwise.

Because [by 23. 6.] fimilar figures are to one another in the duplicate ratio of their homologous fides; the figure described upon BC to the figure defcribed upon BA will be in the duplicate ratio of that of BC to B A. But [by

'1.

. cor. 20, 6.] the fquare of BC to the fquare of B A is in the duplicate ratio of BC to BA: Therefore [by 11. 5.] as the figure defcribed upon BC is to the figure described upon BA, fo is the fquare of BC to the fquare of BA. Alfo by the fame reafon as the figure defcribed upon BC is to the figure defcribed upon CD, fo is the fquare of Be to the fquare of CA: and therefore the figure defcribed upon BC, is to the figures described upon BA, AC together, as the fquare of BC is to the fquares of B A, AC together. But the iquare of BC is [by 47. 1.] equal to the fquares of BA, AC together: Therefore the figure defcribed upon BC is equal to both the figures defcribed upon BA, AC being fimilar to that defcribed upon BC, and having the fame fituation. Which was to be demonftrated.

P This propofition is much more general than the forty-feventh of the first book; it extending to all fimilar figures whatfoever; whereas that is only confined to fquares.

PROP. XXXII. THEOR.

If two triangles having two fides of the one proportional to two fides of the other, be fo fet together at one angle that the homologous fides be parallel: the remaining fides of thofe triangles will be both in one right line.

Let there be two triangles ABC, DCE having two fides BA, AC of the one proportional to two fides CD, DE of the other, viz. as BA is to AC, fo is CD to DE, and let AB be parallel to DC, and AC to DE: I fay BC, CE will be both in the fame right line.

For because AB is paral- A lel to DC, and the right line AC falls upon them; the alternate angles BAC, ACD [by 29. 1.] will be equal to one another. By the fame

D

B

E reason

Book VI. reafon the angle CDE is equal to the angle ACD: Wherefore the angle BA c is equal to the angle cDE And because ABC, DCE are two triangles, having one angle A of the one equal to one angle D of the other, and the fides about the equal angles proportional, viz. as B A is to a c, fo is CD to DE: the triangle ABC [by 6. 6.1 will be equiangular to the triangle DCE: therefore the angle ABC is equal to the angle DCE. But the angle ACD has been proved to be equal to the angle B AC: Therefore the whole angle A CE is equal to both the angles ABC, BAC together. Add the common angle AC B then the angles ACE, ACB together are equal to the angles BAC, ACB, ABC together. But [by 32. 1.] the angles BAC, ACB, ABC together, are equal to two right angles: and therefore the angles ACE, ACB together will be also equal to two right angles. And fo two right lines BC, CE drawn contrariways from the point c in the right line Ac make the adjoining angles ACE, ACB, at the right line A c, equal to two right angles: Therefore [by 14. 1.] BC, CE will both lie in the fame right line.

If therefore two triangles having two fides of the one proportional to two fides of the other, be fo fet together at one angle, that their homologous fides be parallel: the remaining fides of those triangles will be both in the fame right line. Which was to be demonftrated.

PROP. XXXIN. THEOR.

In equal circles the angles have the fame ratio, as the parts of the circumference on which they ftand, whether thofe angles be at the centres, or at the circumferences: and moreover fo are the fectors, as being at the centre.

Let the equal circles be ABC, DEF, and let the angles BGC, EHF be at their centres G, H, and the angles BA C EDF at their circumferences: I fay as the part B C of the circumference is to the part E F, fo is the angle B G C to the angle EHF; the angle BAC to the angle EDF; and moreover fo is the f fector BGC to the fector HEF.

ANE.

For take orderly any number of parts CK, KL of the circumference of one circle each equal to the part B C; and an equal number F M, MN of any other parts of the

circum

circumference of the other circle each equal to EF; and join G K, GL, HM, HN.

A

T

D

Then because the parts BC, CK, KL of the circumference of one circle are equal to one another, the angles BGC, CGK, KGL [by 27. 3.] will be equal to one another: Therefore the part BL of the circumference of one circle is the fame multiple of the part BC of the circumference of that circle, as the angle BGL is of the angle BGC. And by the fame reafon the part E N of the cir cumference of the other circle is the fame multiple of the part EF of the circumference, as the angle EH N is of the angle EHF. And if the part в L of the circumference be equal to the part E N of the other circumference; the angle BGL [by 27. 3.] will be equal to the angle EHN. And if the part BL of the circumference be greater than the part EN; the angle BGL will be greater than the angle EHN: and if lefs, lefs. There are there'fore four magnitudes, viz. the two parts BC, EF of the circumferences of the circles, and the two angles BGC, EHF, and there are taken the equimultiples of the part B C of the circumference of one circle, and of the angle B G C, viz. the part BL of the circumference, and the angle BG L, and equimultiples of the part EF of the circumference of the other circle, and of the angle E H F, viz. the part E N of the circumference, 'and the angle EHN. And it has been demonftrated, that if the part B L of one circumference exceeds the part E N of the circumference of the other circle, the angle BGL alfo exceeds the angle EHN; if PL be equal to E N, the angle BGL is equal to E'HN and if BL be lefs than EN; the angle BG L is lefs than the angle EHN: Therefore [by 5. def. 5.] as the part BC of one circumference is to the part E F of the other, fo is the angle BG C to the angle EHF. But [by 15.5.] as the angle BGC is to the angle EHF, fo is the angle BAC to the angle EDF. For [by 20. 3.] each of them is double to each of thefe; and therefore as the part BC of the one

B

C

H

M

E

F

circum.