Pythagorean Theorem Proof

Of all theorems in mathematics, few have as many proofs as this one — over 370 are known. Below is one of the most elegant: a purely geometric argument requiring no algebra to believe, and very little to verify.

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The TheoremLink to the-theorem

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Proof by Square DissectionLink to proof-by-square-dissection

Construct a large square with side length $a + b$. Its total area is:

$(a + b)^2 = a^2 + 2ab + b^2$

Now place four identical copies of the right triangle inside this square, arranged so their hypotenuses form a tilted inner square. Each triangle has legs $a$, $b$ and hypotenuse $c$, so the inner square has side $c$ and area $c^2$.

The four triangles together have area:

$4 \cdot \frac{1}{2}ab = 2ab$

Since the large square equals the inner square plus the four triangles:

$(a + b)^2 = c^2 + 2ab$

Expanding the left side:

$a^2 + 2ab + b^2 = c^2 + 2ab$

Subtracting $2ab$ from both sides:

$\boxed{a^2 + b^2 = c^2}$

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Proof by Similar TrianglesLink to proof-by-similar-triangles

Draw the altitude from the right angle to the hypotenuse. This splits $\triangle ABC$ into two smaller triangles, both similar to the original.

Label the foot of the altitude $H$. Then:

$\triangle ABC \sim \triangle AHC \sim \triangle CHB$

From the similarity $\triangle ABC \sim \triangle AHC$:

$\frac{b}{c} = \frac{AH}{b} \implies b^2 = c \cdot AH$

From the similarity $\triangle ABC \sim \triangle CHB$:

$\frac{a}{c} = \frac{HB}{a} \implies a^2 = c \cdot HB$

Adding, and using $AH + HB = c$:

$a^2 + b^2 = c \cdot HB + c \cdot AH = c(AH + HB) = c \cdot c$

$\boxed{a^2 + b^2 = c^2}$

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The ConverseLink to the-converse

The converse is equally useful and just as true:

$a^2 + b^2 = c^2 \implies \angle C = 90°$

This gives a purely numerical test for a right angle — no protractor needed. The triples satisfying this are called Pythagorean triples:

Any integer multiple of a Pythagorean triple is also a Pythagorean triple:

$(ka)^2 + (kb)^2 = k^2(a^2 + b^2) = k^2 c^2 = (kc)^2$