Algebraic

The Tilted Square

Traditional · Antiquity · AlgebraicDifficultyOne glance

Expanding the area of a square two different ways — once directly, once as a tilted inner square plus four triangles — forces $a^{2} + b^{2} = c^{2}$ after a single cancellation.

Build a square with side $a + b$ . Going around it, mark a point on each edge a distance $a$ from one corner and $b$ from the next.
Join the four marked points in order. They form a tilted square inside — and each of its sides is a hypotenuse of length $c$ , since each connects the ends of an $a$ and a $b$ meeting at a right angle.
The four corners left over are congruent right triangles, legs $a$ and $b$ , one tucked into each corner of the big square.
The big square's area is the tilted square plus the four triangles: $(a + b)^{2} = c^{2} + 4 (\frac{1}{2} ab)$ .
Expand the left side: $a^{2} + 2 ab + b^{2} = c^{2} + 2 ab$ . The $2 ab$ appears on both sides and cancels outright, leaving $a^{2} + b^{2} = c^{2}$ .

Therefore c² = a² + b². ∎

This is the algebra teacher’s favorite proof: no congruent-triangle argument, no dissection to justify, just one polynomial identity expanded two ways. It is also, not coincidentally, close to the picture behind Bhāskara’s dissection — swap which square holds still and which one tilts, and the two proofs are nearly the same figure.