Recounted in Einstein's autobiographical notes

Any Shape Will Do

Albert Einstein (attributed, age 12) · c. 1891 · Similar trianglesDifficultyClassical

If similar shapes are erected on each side of a right triangle, the shape on the hypotenuse equals the sum of the shapes on the two legs — squares are only the most familiar member of that family.

Let triangle $A B C$ be right-angled at $A$ . This time, forget squares — erect on each side any shape you like, so long as the three shapes are similar to one another.
Drop the altitude from $A$ to $B C$ , meeting it at $D$ . It splits $A B C$ into two smaller triangles, $A B D$ and $A C D$ — each one similar to $A B C$ itself.
So the 'shape' on $A B$ (triangle $A B D$ ) and the shape on $A C$ (triangle $A C D$ ) aren't just similar to $A B C$ — laid out this way, they tile it exactly, with no gap and no overlap.
Any family of similar shapes has area proportional to the square of a chosen reference length: area $= k \cdot (side)^{2}$ , with the same constant $k$ for every member of the family.
Apply that here: $k \cdot A B^{2} + k \cdot A C^{2} = k \cdot B C^{2}$ , since the two pieces make up the whole. Divide out $k$ , and $A B^{2} + A C^{2} = B C^{2}$ — reached without ever drawing a square.

Therefore c² = a² + b². ∎

Einstein describes a proof along these lines in his autobiographical notes, recalling it as one of his early mathematical delights. The core move — that any similar family scales the same way, not just squares — is the same idea that powers The Altitude’s Shadow, pushed one level more general.