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. The construction of squares requires the immediately preceding theorems in Euclid, and depends upon the parallel postulate.
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.
Showing the two congruent triangles of half the area of rectangle BDLK and square BAGF
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 is equal to FB and BD is equal to BC, triangle ABD must be congruent to triangle FBC.
Since A-K-L is a straight line, parallel to BD, then parallelogram BDLK has twice the area of triangle ABD because they share the base BD and have the same altitude BK, i.e., a line normal to their common base, connecting the parallel lines BD and AL. (lemma 2)
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 = AB^2.
Similarly, it can be shown that rectangle CKLE must have the same area as square ACIH = AC^2.
Adding these two results, AB^2 + AC^2 = BD × BK + KL × KC
Since BD = KL, BD* BK + KL × KC = BD(BK + KC) = BD × BC
Therefore AB^2 + AC^2 = BC^2, since CBDE is a square.