If we add to the figure ABIIE, 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

F

G

(AB + BC) (AB – BC) = AB2 – BC2;

which was to be proved.

PROPOSITION ΧΙ. THEOREM.

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 4: then will BC AB2 + AC2.

=

BC equal to BF, for the same reason, and the included angles IIBC 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 BII, and because AC is the prolongation of AL (B. I., P. IV.), their altitudes are equal: hence, the triangle IIBC 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 AII. 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 AII 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,

AB2 = BC2 - AC2 ; or, AC2 = BC2 — AB2.

Cor. 2. If from the vertex of the right angle, a perpendicular be drawn to the hypothenuse, dividing it into twa 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 ВС to BD (P. III.); but the rectangle BE is equal to the square All: hence,

In like manner, we have,

BC: AC2

AC2:: BC : DC.

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

segments of the hypothenuse.

For, by combining the proportions of the preceding corollary

(B. II., P. IV., C.), we have,

B

AB2 AC2 :: BD : DC.

D

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,

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

H D G

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

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

AB2 BC2 =

+402 - 2BC × CD.

B

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.),

BD2 BC2 + CD2 2BC x CD.

=

Adding AD2

to both members, we

have,

BD2 + AD2 = BC2 + CD2 + AD2 — 2BC × CD.

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 rect angle 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 the perpendicular drawn. from A to BC produced; then will

AC2 = BC2 + AB2 + 2BC × BD.

For, CD is the sum of BC

and BD hence (P. VIII.),

=

CD2 BQ2

+ BD2 + 2BC × BD.

Adding AD to both members,

and reducing, we have,

2

B

AC2 = BO2 + AB2 + 2BC × BD;

which was to be proved.

Scholium. The right-angled triangle is the only one which the suni 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