"Behold!" is the traditional caption for this image, but that's just a snotty
mathemetician's way of saying "I worked out this algebra before you did!"

So: let the length of the sides of the whole image be A. The image is square. Let the side length of the red square be B. The blue triangle is a right triangle. It has sides of length A (on the outside), C (shortest), and D.

Cleverly, this diagram is drawn so that D = B + C. Any right triangle can be have this diagram drawn about it. The Pythagorean theorem, for the blue triangle, is AxA = CxC + DxD.

The area of the big square is AxA. The area is also the area of the red square plus 4 of the blue triangles. The area of a triangle is .5x base x height. For a right triangle, you can use either side touching the right angle as the height. So:

AxA = BxB + 4x.5(CxD)

AxA = BxB + 2(CxD)

AxA = (D-C)x(D-C) + 2xCxD

AxA = DxD - 2xCxD + CxC + 2xCxD

AxA = CxC + DxD

"Behold", indeed.

Please look at the source code of this page for more information.

So: let the length of the sides of the whole image be A. The image is square. Let the side length of the red square be B. The blue triangle is a right triangle. It has sides of length A (on the outside), C (shortest), and D.

Cleverly, this diagram is drawn so that D = B + C. Any right triangle can be have this diagram drawn about it. The Pythagorean theorem, for the blue triangle, is AxA = CxC + DxD.

The area of the big square is AxA. The area is also the area of the red square plus 4 of the blue triangles. The area of a triangle is .5x base x height. For a right triangle, you can use either side touching the right angle as the height. So:

AxA = BxB + 4x.5(CxD)

AxA = BxB + 2(CxD)

AxA = (D-C)x(D-C) + 2xCxD

AxA = DxD - 2xCxD + CxC + 2xCxD

AxA = CxC + DxD

"Behold", indeed.

Please look at the source code of this page for more information.