An indirect algebraic proof, said to be due to the great Leibniz (1646-1716).
If (1) HA² + HB²= AB²,
then (2) HA² = AB² - HB²,
whence (3) HA² = (AB + HB)
(AB - HB).
Take BE and BC each equal to AB, and from B as center describe the semicircle CA'E. Join AE and AC, and draw BD perp. to AE. Now (4) HE = AB + HB, and (5) HC = AB - HB. (4) × (5) gives HE × HC = HA², which is true only when triangles AHC and EHA are similar.
∴ (6) angle CAH = angle AEH, and so (7) HC
: HA = HA : HE; since angle HAC = angle E, then angle CAH = angle EAH. ∴ angle AEH + angle EAH = 90° and angle CAH + angle EAH = 90°. ∴ angle EAC = 90°. ∴ vertex A lies on the semicircle, or A coincides with A'. ∴ EAC is inscribed in a semicircle and is a rt. angle. Since equation (1) leads through the data derived from it to a rt triangle, then starting with such a triangle and reversing the argument we arrive at h² = a² + b².
a. See Versluys, p. 61, fig. 65, as given by von Leibniz.