Pythagoras' theorem

It is the most famous theorem in elementary geometry. We prove it by equivalence of areas, exploiting the idea of equidecomposability.

Theorem — Pythagoras'

In a right-angled triangle, the square built on the hypotenuse is equivalent to the sum of the squares built on the legs. $\begin{cases} AC \perp BC \\ ABHI,\; AEDC,\; CBFG \text{ quadrati} \end{cases} \implies ABHI \doteq AEDC + CBFG$

Proof

The height $CK$, extended to $M$, divides the square on the hypotenuse into two rectangles, each equivalent to one of the squares built on the legs.

1. Draw $CK\perp AB$ and extend it to the side $HI$ of the square on the hypotenuse (point $M$). The line $KM$ divides the large square into two rectangles.
2. One checks that the rectangle $AKMI$ is equivalent to the square $AEDC$ (built on the leg $AC$): both have the same base and the same height.
3. By the same reasoning, the rectangle $KBHM$ is equivalent to the square $CBFG$ (built on the leg $BC$).
4. Adding the two rectangles reassembles the square on the hypotenuse: $AKMI + KBHM = ABHI$, hence $c^2 = a^2+b^2$.

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Check the theorem dynamically: drag the triangle’s vertices and compare the areas of the squares built on the sides.

Drag the vertices $A$ and $B$: together the green squares on the legs always match the red square on the hypotenuse.

Links

Topics: Equivalence and Pythagoras
Concepts: Area · Equidecomposability · Equivalence · Square · Pythagoras’ theorem · Right-angled triangle
Methods: Equivalence of figures · Pythagoras
Skills: Proving · Synthetic geometry
People: Pythagoras