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Book II.

PROP. XIII,

THEOR.

I

N every triangle, the square of the side subtending

the fides containing' that angle, by twice the rectangle contained by either of these fides, 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 sides containing it,

let fall the perpendiculare AD from the opposite angle : The a 12. 3. square of AC, opposite 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 because the straight line CB is di. vided into two parts in the point Db, BC? +BD2 = 2BC.BD+CD2.

b 7. 2. Add to each AD? ; then BC? + BDP + AD = 2BC.BD + CD? + AD But BDP + AD AB, and CDP+DA? = AC?c; therefore BCP+AB2=2BC.BD +AC?;

C 47. 1. that is, ACis less than BC?+ B. D AB? by 2BC.BD.

Secondly, Let AD fall without the triangle ABC *: Then because the angle at D is a right angle, the angle ACB is greaterd than a right angle, and AB = AC?+BC2+2BC.CD. d. 16. I. Add BC to each ; then AB+BC=AC2+2BC2+2 BC.CD.

But

CI2. 2.

+ See figure of the last Fropofition.

Book II. But because BD is divided into two parts in C, BC? + f3 2.

BC.CD= BC.BD, and 2BCP+2BC.CD=2BC.BD : therefore ABP+BC = AC +2BC.BD; and AC is less than ABP+BC?, by 2BD.BC.

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 manifest that 8 47. 1. & AB? + BC = AC? + 2BC2 = AC? +

2BC.BC Therefore in every triangle,
&c. Q. E. D.

B

PROP. XIV. PROB.

.

o describe a square that shall be equal to a

given rectilineal .

a 45. I.

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

Describe a the rectangular parallelogram BCDE equal to the rectilineal figure A. If then the sides 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 ; tbere

fore

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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 are equal c to the square of GH:

H Therefore also the rectangle BE.EF А together with the square of EG, is equal

the squares of HE

G and EG. Take away the square

C

D of EG, which is 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 described

upon

EH. Which was to be done.

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F one side of a triangle be bisected, the sum of the See N.

squares of the other two fides is double of the square of half the side bilected, 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.

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Book II. From A draw AE perpendicular to BC, and because

BEA is a right angle, AB2=a a 47. I.

A
BE? + AE?, and AC = CE?+
AE?; wherefore AB++AC
BE?+CE+2AE?. But because

the line BC is cut equally in D, b9 2. andunequally in E,BE?+CE=)

2 BD-+2DE?; therefore AB+
AC = 2BD? + 2DE + 2 AE2,
Now DE? + AE = ADP, and B
2DE? + 2 AE = 2AD-; wherefore ABP + ACP = 2BD? +
2 AD. Therefore, &c. Q. E. D.

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HE sum of the squares of the diameters of any

T is

of the sides 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 fquares of AB, BC, CD, DA.

Let AC and BD intersect one another in E: and because a 15.1.

the vertical angles AED, CEB are equal a, and also the al. b 29. 1.

ternate angles EAD, ECB b, the triangles ADE, CEB have two angles in the one equal to two angles in the other, each to each : but the sides AD and BC, which are opposite to equal

angles in these triangles, C 34. I. are also equal c; there. fore the other fides which

A

D are oppofite to the equal d 26. 1. angles are alfo equal,

E
viz. AE to EC, and ED
to EB.

Since; therefore, BD is
bifected in E, AB+B

С
AD2BE2+2 AE ; and
e A. 2.

for the same reason, CD’+
BCP = 2BE+ 2EC2 = 2BE? + 2 AE?, because EC=AE.
Therefore. AB + ADP+DCP + BC = 4BE? + 4AE?. But

4BE

BE?=BD, and 4AE-AC2f, because BD and AC are Book II. both bisected in E; therefore AB?+ADP+CD?+BC2=

f2. Cor. 8.2. BD2+AC?. Therefore the sum of the squares, &c. Q. E. D.

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

E LE

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