The truth of this proposition has been proved geometrically in Art. 353. It will now be shown graphically and proved algebraically.
Let A BCD (Fig. 284) be a rectangle of equal sides, and BED the right-angled triangle, the squares upon the sides of which, it is proposed to consider. Extend the side BE to F; parallel with BF draw DG, CK, and A L. Parallel with ED draw A J and L G. These lines produce triangles, AHB, AC J, ALC, CKD, and CGD, each equal to the given triangle BED (Art. 337). Now, if from the square A B CD we take A B H and place it at CD G; and if we take BED and place it at A L C we will modify the square A BCD, so as to produce the figure LGDEHA L, which is made up of two squares, namely, the square D E FG and the square ALFH, and these two squares are evidently equal to the square A B CD. Now, the square DE FG is the square upon ED, the base of the given right-angled triangle, and the square A L FH is the square upon A H = B E, the perpendicular of the given right-angled triangle, while the square A B CD is the square upon BD, the hypothenuse of the given right-angled triangle. Thus, graphically, it is shown that the square upon the hypothenuse of a right-angled triangle is equal to the sum of the squares upon the remaining two sides. To show this algebraically, let B E, the perpendicular of the given right-angled triangle, be represented by a; E D, the base, by b, and B D, the hypothenuse, by c. Then it is required to show that -
c2 = a2 + b2 .
Now, since D K = B E = a, therefore, E K = E D - D K = b - a, and the square E KJ H equals (b - a2), which (Art. 413) equals
b2 - 2ab + a2
This is the value of the square E K JH which, with the four triangles surrounding it, make up the area of the square A BC D. Placing the triangle A B H of this square outside of it at CD G, and the triangle BE D at A L C, we have the four triangles, grouped two and two, and thus forming the two rectangles C G D K and A LCJ. Each of these rectangles has its shorter side (A L, CG) equal to BE - a, and its longer side L C, G D, equal to E D = b; and the sum of the two rectangles is ab + ab=2ab. This represents the area of the two rectangles, which are equal to the four triangles, which, together with the square EKJH, equal the square A BCD; or -
or - c2 = (b - a)2 + a b + a b, or -
c2 = (b - a)2 + 2ab.
Then, substituting for (b - of, its equivalent as above, we have -
c2 = b2 - 2ab + a2+2ab.
Remove the two like quantities with unlike signs (Art. 402), and we have -
which was to be proved.