Comment by russfink

May be a repeat here, but best proof I saw was inscribe a square with sides of length c inside another square, but rotated such that the interior square’s corners intersect the outer square’s edges. The intersecting points divide the outer square’s edge making lengths a and b.

This produces an inner square’s edge with sides length c and four equal right triangles of sides a, b, and c.

Note that the area of the outer square equals the sum of the inner square plus the area of the four triangles. Solve this equality.