9.6 An Extra Proof: Rearranging Four Triangles

The combining construction in 9.2 already proves the theorem. Here is a second, self-contained proof that many students find especially convincing, and it needs nothing but areas.

Take any right triangle with legs a and b and hypotenuse c. Make four identical copies, and pack them into a large square of side a + b in two different ways.

Arrangement 1. Place the four triangles in the corners so that a tilted square of side c is left empty in the middle. The big square’s area splits as (a+b)^2 = 4 \times \tfrac12 ab + c^2.

Arrangement 2. Slide the same four triangles to leave two square gaps instead, one of side a, one of side b. Now (a+b)^2 = 4 \times \tfrac12 ab + a^2 + b^2.

Both describe the same big square, so the leftover regions must match: c^2 = a^2 + b^2.

The four triangles cancel out of both pictures, and the bare equality of the empty space is the theorem itself. It is hard to imagine a cleaner argument.

Try This Cut four congruent right triangles from card (try legs 5 and 12). Arrange them both ways inside a square of side 5 + 12 = 17. Confirm by eye that the empty middle in Arrangement 1 is a square, and that its area 13^2 = 169 equals 5^2 + 12^2 = 25 + 144 = 169 from Arrangement 2.

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