*

*

For draw CE from the Point C at Right Angles 11. 1. to AB, which make equal to AC, or CB, and draw AE, EB; likewife thro' E let EF be + drawn parallel † 31. 1. to AD, and thro D, DF + parallel to CE.

Then because the Right Line EF falls upon the Parallels EC, FD, the Angles CEF, EFD, are equal ‡ 29. 1. to two Right Angles. Therefore the Angles FEB, EFD, are together less than two Right Angles. But Right Lines making, with a third Line, Angles to→ gether less than two Right Angles, being_infinitely produced, will meet *. Wherefore EB, FD, pro- * Ax. 12. duced, will meet towards BD. Now let them be produced, and meet each other in the Point G, and let AG be drawn.

And then becaufe AC is equal to CE, the Angle AEC will be equal to the Angle EAC+: But the † 5. 1. Angle at C is a Right Angle. Therefore the Angle CAE, or AEC, is half a Right one. By the fame way of Reasoning, the Angle CEB, or EBC, is half a Right one. Therefore AEB is a Right Angle, And fince EBC is half a Right Angle, DBG will falfo ‡ 15. 1. be half a Right Angle, fince it is vertical to CBE. But BDG is a Right Angle alfo; for it is * equal to * 29. 1. the alternate Angle DCE. Therefore the remaining Angle DGB is half a Right Angle, and fo equal to DBG. Wherefore the Side BD is † equal to the † 6. 1. Side DG. Again, because EGF is half a Right Angle, and the Angle at F is a Right Angle, for it is equal to the oppofite Angle at C; the remaining Angle FEG will be alfo half a Right one, and is equal to the Angle EGF; and fo the Side GF is † equal to the Side EF. And fince EC is equal to CA, and the Square of EC equal to the Square of CA; there fore the Squares of EC, CA, together, are double to the Square of CA. But the Square of EA is equal ‡ 47. to the Squares of EC, CA. Wherefore the Square of EA is double to the Square of AC. Again, becaufe GF is equal to FE, the Square of GF alfo is equal to the Square of FE. Wherefore the Squares of GF, FE, are double to the Square of F E. But the Square of EG ist equal to the Squares of GF, FE. Therefore the Square of EG is double to the Square of EF: But EF is equal to CD. Wherefore the Square of EG fhall be double to the Square of CD.

E 4

But

+ 47. I

46. I.

But the Square of EA has been proved to be double to the Square of AC. Therefore the Squares of AE, EG, are double the Squares of AC, CD. But the Square of AG ist equal to the Squares of AE, EG; and confequently the Square of AG is double to the Squares of AC, CD. But the Squares of AD, DG, are equal to the Square AG. Therefore the Squares of AD, DG, are double the Squares of AC, CD. But DG is equal to DB. Wherefore the Squares of AD, DB, are double to the Squares of AC, CD. Therefore if a Right Line be cut into two equal Parts, and to it be directly added another; the Square made on [the Line compounded of] the whole Line, and the added one, together with the Square of the added Line, fhall be double to the Square of the half Line, and the Square of [that Line which is compounded of] the half, and the added Line.

To cut a given Right Line fo, that the Rectangle contained under the whole Line, and one Segment, be equal to the Square of the other Segment.

L

ET AB be a given Right Line. It is required

to cut the fame fo, that the Rectangle contained under the whole, and one Segment thereof, be equal to the Square of the other Segment.

Defcribe* ABCD the Square of AB, bifect A C in E, and draw BE: Alfo, produce CA to F, fo that EF be equal to EB. Defcribe F G HA the Square of AF, and produce G H to K. I fay, AB is cut in H fo, that the Rectangle under AB, BH, is equal to the Square of A H.

r

For fince the Right Line A C is bifected in E, and AF is directly added thereto, the Rectangle under CF, FA, together with the Square of AE, will be +6 of this. + equal to the Square of EF. But EF is equal to EB. Therefore the Rectangle under CF, FA, together with the Square of AE, is equal to the Square of EB. But the Squares of BA, AE, are equal to the Square of EB; for the Angle at A is a Right An

$ 47. I.

gle.

gle. Therefore the Rectangle under CF, FA, together with the Square of AE, is equal to the Squares of BA, AE. And taking away the Square of AE, which is common, the remaining Rectangle under CF, FA, is equal to the Square of A B. But FK is the Rectangle under CF, FA; fince AF is equal to FG; and the Square of AB is AD. Wherefore the Rectangle FK is equal to the Square AD. And if AK, which is common, be taken from both, then the remaining Square FH is equal to the remaining Rectangle HD. But HD is the Rectangle under AB, BH, fince AB is equal to BD, and FH is the Square of AH. Therefore the Rectangle under AB, BH fhall be equal to the Square of AH. And fo the given Right Line AB is cut in H, so that the Rectangle under AB, BH, is equal to the Square of AH; which was to be done.

PROPOSITION XII.

THIORE M.

In obtufe angled Triangles, the Square of the Side fubtending the obtufe Angle, is greater than the Squares of the Sides containing the obtufe Angle, by twice the Rectangle under one of the Sides, containing the obtufe Angle, viz. that on which, produced, the Perpendicular falls, and the Line taken without, between the perpendicular and the obtufe Angle.

ET ABC be an obtufe angled Triangle, having the obtufe Angle BAC; and from the 12. » Point B draw BD perpendicular to the Side CA produced. I fay the Square of BC is greater than the Squares of BA and AC, by twice the Rectangle contained under CA, and AD.

For because the Right Line CD is any how cut in the Point A, the Square of CD fhall be † equal to the t 4 of this. Squares of CA, and AD, together with twice the Rectangle under CA, and AD. And if the Square of BD, which is common, be added, then the Squares of CD, DB, are equal to the Squares of CA, AD,

47. 1.

12. I.

*

and DB, and twice the Rectangle contained under CA and AD. But the Square of CB is equal to the Squares of CD, DB; for the Angle at D is a Right one, fince BD is perpendicular, and the Square of AB is equal to the Squares of AD, and DB: Therefore the Square of CB is equal to the Squares of CA and AB, together with twice a Rectangle under CA and AD. Therefore in obtufe angled Triangles, the Square of the Side fubtending the obtufe Angle, is greater than the Squares of the Sides containing the obtufe Angle, by twice the Rectangle under one of the Sides containing the obtufe Angle, viz. that on which, produced, the Perpendicular falls, and the Line taken without, between the perpendicular and the obtufe Angle; which was to be demonftrated.

PROPOSITION XIII.

THEOREM II.

In acute angled Triangles, the Square of the Side fubtending the acute Angles, is less than the Squares of the Sides containing the acute Angle, by twice a Rectangle under one of the Sides about the acute Angle, viz. on which the Perpendicular falls, and the Line affumed within the Triangle, from the Perpendicular to the acute Angle.

ET ABC be an acute angled Triangle, having

drawn AD perpendicular to BC. I fay the Square of AC is lefs than the Squares of CB and BA by twice a Rectangle under CB and BD.

For because the Right Line CB is cut any how in +7 of this. D, the Squares of CB and BD will be + equal to twice a Rectangle under CB and BD, together with the Square of DC. And if the Square of AD be added to both, then the Squares of CB, BD, and DA, are equal to twice the Rectangle contained under CB and BD, together with the Squares of AD and DC. But the Square of AB is equal to the Squares of BD and DA; for the Angle at D is a Right Angle. And the Square of AC is equal to

the

the Squares of AD and DC. Therefore the Squares of CB and BA are equal to the Square of A C, together, with twice the Rectangle contained under CB and BD. Wherefore the Square of AC only, is lefs than the Squares of CB and BA, by twice the Rectangle under CB and BD. Therefore in acute angled Triangles, the Square of the Side fubtending the acute Angles, is less than the Squares of the Sides containing the acute Angle, by twice a Rectangle under one of the Sides about the acute Angle, viz. on which the Perpendicular falls, and the Line affumed within the Triangle, from the perpendicular to the acute Angle; which was to be demonftrated.

PROPOSITION XIV.

PROBLEM.

To make a Square equal to a given Right-lined
Figure.

ET A be the given Right-lined Figure. It is
required to make a Square equal thereto.

Make the Right-angled Parallelogram BCDE* 45. 1. equal to the Right-lined Figure A. Now if BE be equal to ED, what was propofed will be already done, fince the Square BD is made equal to the Rightlined Figure A: But if it be not, let either BE or ED be the greater: Suppofe BE, which let be produced to F; fo that EF be equal to ED. This being done, let BE be + bifected in G, about which, as a Center, † 10. 1. with the Distance GB or GF, defcribe the Semicircle BHF; and let DE be produced to H, and draw GH. Now because the Right Line BF is divided into two equal Parts in G, and into two unequal ones in E, the Rectangle under BE and EF, together with the Square of GE, fhall be equal to the Square of GF. 15 of this. But GF is equal to GH. Therefore the Rectangle under BE, EF, together, with the Square of GE, is equal to the Square of GH. But the Squares of HE and EG are equal to the Square of GH. 47. I. Wherefore the Rectangle under BE, EF, together with the Square of EG, is equal to the Squares of HE, EG, And if the Square of EG, which is com

mon