Page:The Elements of Euclid for the Use of Schools and Colleges - 1872.djvu/75

Rh Now the parallelogram BL is double of the triangle ABD, because they are on the same base BD, and between the same parallels BD, AL. [I. 41. And the square GB is double of the triangle FBC, because they are on the same base FB, and between the same parallels FB, GC. [I. 41. But the doubles of equals are equal to one another. [Ax. 6. Therefore the parallelogram BL is equal to the square GB.

In the same manner, by joining AE, BK, it can be shewn, that the parallelogram CL is equal to the square CH. Therefore the whole square BDEC is equal to the two squares GB, HC. [Axiom 2. And the square BDEC is described on BC, and the squares GB,HC on BA,AC Therefore the square described on the side BC is equal to the squares described on the sides BA, AC.

Wherefore, in any right-angled triangle &c.

PROPOSITION 48. THEOREM.

If the square described on one of the sides of a tri-angle be equal to the squares described on the other two sides of it, the angle contained by these two sides is a right angle.

Let the square described on BC, one of the sides of the triangle ABC, be equal to the squares described on the other sides BA, AC: the angle BAC shall be a right angle.

From the point A draw AD at right angles to AC ; [I. 11. and make AD equal to BA; [I. 3. and join DC.

Then because DA is equal to BA, the square on DA is equal to the square on BA. To each of these add the square on AC. Therefore the squares on DA, AC are equal to the squares on BA,AC [Axiom 2.