PROP. XIII. THEOR.

Book II.

N every triangle, the square of the fide fubtending any of the acute angles, is less than the squares of the fides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular, let fall upon it from the opposite angle, and the acute angle.

Let ABC be any triangle, and the angle at B one of its acute angles, and upon BC, one of the fides containing it, let fall the perpendicular AD from the opposite angle: The a 12.1. square of AC, oppofite to the angle B, is less than the squares of CB, BA by twice the rectangle CB.BD.

First, Let AD fall within the triangle ABC; and be

cause the straight line CB is di

vided into two parts in the point

Db, BC2+BD2 = 2BC.BD+CD2.

b 7.2.

C 47.1.

that is, AC2 is less than BC2+ B D

C

AB2 by 2BC.BD.

Secondly, Let AD fall without the triangle ABC*: Then

because the angle at Dis a right angle, the angle ACB is great

erd than a right angle, and AB

2

AC2+BC2+2BC.CD. d. 16. 1.

Add BC2 to each; then AB2+BC2 AC2+2BC2+2BC.CD.

C 12. 2.

But

1

Book II.

f3 2.

But because BD is divided into two parts in C, BC2 +
BC.CDf BC.BD, and 2BC2+2BC.CD=2BC.BD: there-
fore AB2+BC2 = AC2+2BC.BD; and AC2 is less than
AB2+BC2, by 2BD.BC.

A

Lastly, Let the fide AC be perpendicular to BC; then is BC the straight line between the perpendicular and the acute angle at B; and it is manifeft that 8 47. 1. g AB2 + BC2 = AC2 + 2BC2 = AC2 + 2BC.BC. Therefore in every triangle, &c. Q. E. D.

Teringat

O describe a square that shall be equal to a

Let A be the given rectilineal figure; it is required to describe a square that shall be equal to A.

Describe a the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the fides of it, BE, ED are equal to one another, it is a square, and what was required is done; but if they are not equal, produce one of them BE to F, and make EF equal to ED, and bisect BF in G: and from the centre G, at the distance GB, or GF, describe the femicircle BHF, and produce DE to H, and join GH. Therefore, because the straight line BF is divided into two equal parts in the point G, and into two unequal in the point E, the rectangle BE.EF, together with the square of EG, is equal to the square of GF: but GF is equal to GH; there

fore the rectangle BE.EF, together with the square of EG, Book II. is equal to the square of GH: But the squares of HE and EG

common to both, and the remaining rectangle BE.EF is equal to the square of EH: But BD is the rectangle contained by BE and EF, because EF is equal to ED; therefore BD is equal to the square of EH; and BD is also equal to the rectilineal figure A; therefore the rectilineal figure A is equal to the square of EH: Wherefore a square has been made equal to the given rectilineal figure A, viz. the square defcribed upon EH. Which was to be done.

PROP. A. THEOR.

I

F one fide of a triangle be bisected, the fum of the See N. squares of the other two fides is double of the square of half the fide bisected, and of the square of the line drawn from the point of bisection to the opposite angle of the triangle.

Let ABC be a triangle, of which the fide BC is bisected in D, and DA drawn to the opposite angle; the squares of BA and AC are together double of the squares of BD and DA.

1

Book II.

يا

a 47. I.

From A draw AE perpendicular to BC, and because

BEA is a right angle, AB2= a
BE2 + AE2, and ACCE2+
AE2; wherefore AB2+AC
BE2+CE2+2AE2. But because

the line BC is cut equally in D,
b9 2. andunequally in E, BE2+CE2_b
2BD+2DE2; therefore AΒ2+
AC22BD2 + 2DE2 + 2AE2.
Now DE2 + AE = AD2, and B

D

A

EC

2DE2 + 2AE2 = 2AD2; wherefore AB2 + AC2 = 2BD2 +

2AD2. Therefore, &c. Q. E. D.

a 15. 1.

b 29. 1.

PROP. В. THEOR.

HE fum of the squares of the diameters of any

I parallelogram is equal to the fum of the squares

of the fides of the parallelogram.

Let ABCD be a parallelogram, of which the diameters are AC and BD; the sum of the squares of AC and BD is equal to the sum of the squares of AB, BC, CD, DA,

Let AC and BD interfect one another in E: and because the vertical angles AED, CEB are equal a, and also the alternate angles EAD, ECBb, the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each: but the fides AD and BC, which are opposite to equal angles in these triangles,

34. 1. are alfo equal; therefore the other fides which are oppofite to the equal d 26. 1. angles are alfo equal d, viz. AE to EC, and ED to EB.

A

D

E

C

Since, therefore, BD is bisected in E, AB2 + B A. 2. AD2BE2+2AE2; and for the fame reason, CD2+ BC22BE2 + 2EC22BE2 + 2AE, because EC AE. Therefore AB2 + AD2+DC2 + BC24BE2 + 4AE2. But

1

4BE

4BE2BD2, and 4AEAC2 f, because BD and AC are Book II. both bisected in E; therefore AB2+AD2+CD2+BC2= f 2. Cor. 8.2. BD2+AC2. Therefore the sum of the squares, &c. Q. E.

D.

Cor. From this demonstration, it is manifest that the diameters of every parallelogram bisect one another.