147. 1.

46. 1.

But the Square of EA has been prov'd to be double to the Square of AC. Therefore the Squares of AE, EG, are double the Squares of AC, CD. But the Squares of A G is 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,

PROPOSITION. XI.

PROBLEM.

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.

LET 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 AC in E, and draw BE: Alfo, preduce CA to F, fo that E F be equal to E B. Defcribe F G H A the Square of A F, and produce GH to K. I fay, AB is cut in H fo, that the Rectangle under AB, BH, is equal to the Square of AH.

For fince the Right Line AC is bifected in E, and A E is directly added thereto, the Rectangle under CF, FA, together with the Square of AE, will be +6of 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 $47 of this. of EB. But the Squares of BA, AE, are equal to the Square of EB; for the Angle at A is a Right An

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 AB. But F K is the Rectangle under CF, FA; fince AF is equal to FG; and the Square of AB is AD. Wherefore the Rectangle F K is equal to the Square AD. And if A K, 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 ÁH. Therefore the Rectangle under A B, BH fhall be equal to the Square of A H; And fo the given Right Line AB is cut in H, fo that the Rectangle under AB, BH, is equal to the Square of AH. Which was to be done.

PROPOSITION XII.

THEOREM.

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, produc'd, the Perpendicular falls, and the Line taken without, between the perpendicular and the obtuse Angle.

L

ET ABC be an obtufe angled Triangle, having the obtufe Angle BAC; and * from the Point B * 12. 1, draw BD perpendicular to the Side CA produc'd. I fay the Square of BC is greater than the Squares of BA and AC, by twice the Rectangle contain❜d 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 † 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, and DB, and twice the Rectangle contain❜d under

* 47 I.

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, produc'd, the Perpendicular falls, and the Line taken without, between the perpendicular and the obtufe Angle; which was to be demonftrated."

PROPOSITION XIII.

THEOREM.

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 affum'd within the Triangle, from the Perpendicular to the acute Augle.

ET ABC be an acute angled Triangle, having

*

LE
the acute Angle B: And from A let there be
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 contain❜d under CB and BD, together with the Squares of AD #47.1. 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 Squares of AD and DC. Therefore the Squares of CB and BA are equal to the Square of AC, to

gether

gether, with twice the Rectangle contain'd 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 affum'd within the Triangle, from the perpendicular to the acute Angle; which was to be demonstrated.

To make a Square equal to a given Right-lin❜d Figure. LETA be the given Right-lin❜d Figure. It is requir'd to make a Square equal thereto.

Make the Right-angled Parallelogram BCDE 45. §. equal to the Right-lin'd Figure A. Now if BE be equal to ED, what was propos'd will be already done, fince the Square BD is made equal to the Rightlin'd Figure A: But if it be not, let either BE or ED be the greater: Suppofe BE, which let be produc'd to F; fo that EF be equal to ED. This being done, let BE be + bifected in G, about which, as a Center, † 1o. 1. with the Distance GB or GF, defcribe the Semicircle BHF; and let DE be produc'd 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 G E, fhall be ‡ equal to the Square of GF. +5 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. 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 common, be taken from both, the remaining Rectangle contain'd under BE and EF, is equal to the Square of EH. But the Rectangle under BE and EF is the Parallelogram B D, becaufe EF is equal to ED.

There

*

47.1.

Therefore the Parallelogram BD is equal to the Square of EH; but the Parallelogram BD is equal to the Right-lin❜d Figure A. Wherefore the Rightlin'd Figure A is equal to the Square of EH. And fo there is a Square made equal to the given Rightlin'd Figure A, viz. the Square of EH; which was to be done.

The END of the SECOND BOOK.

EUCLID'S