The proof is as follows:

Let ACB be a right-angled triangle with right angle CAB.
On each of the sides BC, AB, and CA, squares are drawn, CBDE, BAGF, and ACIH, in that order.
From A, draw a line parallel to BD and CE. It will perpendicularly intersect BC and DE at K and L, respectively.
Join CF and AD, to form the triangles BCF and BDA.
Angles CAB and BAG are both right angles; therefore C, A, and G are collinear. Similarly for B, A, and H.
Angles CBD and FBA are both right angles; therefore angle ABD equals angle FBC, since both are the sum of a right angle and angle ABC.
Since AB and BD are equal to FB and BC, respectively, triangle ABD must be equal to triangle FBC.
Since A is collinear with K and L, rectangle BDLK must be twice in area to triangle ABD.
Since C is collinear with A and G, square BAGF must be twice in area to triangle FBC.
Therefore rectangle BDLK must have the same area as square BAGF = AB2.
Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC2.
Adding these two results, AB2 + AC2 = BD × BK + KL × KC
Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC
Therefore AB2 + AC2 = BC2, since CBDE is a square.