If we add to the figure ABHE, the rectangle BKLH, we shall have the rectangle AKLE, which is equal to the the rectangle of AB + BC and AB BC. If to the same figure

АВНЕ, We add the rectangle

DGFE, equal to equal to

BKLH,

F

G Ι

H

we

E

D

The square described on the hypothenuse of a right-angled triangle, is equal to the sum of the squares described on the other two sides.

Let ABC be a triangle, right-angled at A: then will B02 = AB2 + AC2.

2

BC equal to BF, for the same reason, and the included angles HBC and ABF equal, because each is equal to the angle ABC plus a right angle hence, the triangles are equal in all their parts (B. I., P. V.).

The triangle ABF, and the rectangle BE, have the same base BF, and because DE is the prolongation of DA, their altitudes are equal: hence, the triangle ABF is equal to half the rectangle BE (P. II.). The triangle HBC, and the square BL, have the same base BH, and because AC AC is the prolongation of AL (B. I., P. IV.), their altitudes are equal: hence, the triangle HBC is equal to half the square of AH. But, the triangles ABF and HBC are equal: hence, the rectangle BE is equal to the square AH. In the same manner, it may be shown that the rectangle DG is equal to the square AI: hence, the sum of the rectangles BE and DG, or the square BG, is equal to the sum of the squares AH and AI; or, BC2 AB2 + AC2; which was to be proved.

Cor. 1. The square of either side about the right angle is equal to the square of the hypothenuse diminished by the square of the other side: thus,

ᎪᏴ = - ;
= BC2 — AC2; or, A C2

BC2 - AB2.

Ᏼ

right angle, a per

Cor. 2. If from the vertex of the pendicular be drawn to the hypothenuse, dividing it into two segments, BD and DC, the square of the hypothenuse will be to the square of either of the other sides, as the hypothenuse is to the segment adjacent to that side.

For, the square BG, is to the rectangle BE, as BC to BD (P. III.); but the rectangle BE is equal to the square AH: hence,

Cor. 3. The squares of the sides about the right angle are to cach other as the adjacent

Cor. 4. The square described on the diagonal of a square is double the given square.

For, the square of the diagonal is equal to the sum of the squares of the two sides; but the square of each side is equal to the given square: hence,

H

D

G

A

E B F

AC2

2AB2 ; or, AC2
C2 2BC2.

Cor. 5. From the last corollary, we have,

ÃÛ2 : AB2 : 2 : 1;

hence, by extracting the square root of each term, we have,

AC : AB : √2 : 1;

that is, the diagonal of a square is to the side, as the square root of two to one; consequently, the diagonal and the side of a square are incommensurable.

PROPOSITION XII. THEOREM.

In any triangle, the square of a side opposite an acute angle, is equal to the sum of the squares of the base and the other side, diminished by twice the rectangle of the base and the distance from the vertex of the acute angle to the foot of the perpendicular drawn from the vertex of the opposite angle to the base, or to the base produced.

Let ABC be a triangle, C one of its acute angles, BC its base, and AD the perpendicular drawn from A to BC, or BC produced; then will

A

For, whether the perpendicular meets the base, or the base produced, we have BD equal to the difference of BC and CD: hence (P. IX.),

In any obtuse-angled triangle, the square of the side opposite the obtuse angle is equal to the sum of the squares of the base and the other side, increased by twice the rectangle of the base and the distance from the vertex of the obtuse angle to the foot of the perpendicular drawn from the vertex of the opposite angle to the base produced.

Let ABC be an obtuse-angled triangle, B its obtuse angle, BC its base, and AD base, and AD the perpendicular drawn from A to BC produced; then will

Adding AD to both members,

and reducing, we have,

D B

AC2 – BC2 + AB2 + 2BC × BD ;

which was to be proved.

Scholium. The right-angled triangle is the only one in which the sum of the squares described on two sides is equal to the square described on the third side.

PROPOSITION XIV. THEOREM.

In any triangle, the sum of the squares described on two sides is equal to twice the square of half the third side, increased by twice the square of the line drawn from the middle point of that side to the vertex of the opposite angle.

Let ABC be any triangle, and EA a line drawn from