A Proof of the Pythagorean Theorem From Heron's Formula

Let the sides of a triangle have lengths a,b and c. Introduce the semiperimeter p = (a + b + c)/2 and the area S. Then Heron's formula asserts that

S² = p(p - a)(p - b)(p - c)

W. Dunham analyzes the original Heron's proof in his Journey through Genius.

For the right triangle with hypotenuse c, we have S = ab/2. We'll modify the right hand side of the formula by noting that

p - a = (- a + b + c)/2,
p - b = (a - b + c)/2,
p - c = (a + b - c)/2.

It takes a little algebra to show that

16S² = (a + b + c)(- a + b + c)(a - b + c)(a + b - c)

= 2a²b² + 2a²c² + 2b²c² - (a⁴ + b⁴ + c⁴)

For the right triangle, 16S² = 4a²b². So we have

4a²b²= 2a²b² + 2a²c² + 2b²c² - (a⁴ + b⁴ + c⁴)

Taking all terms to the left side and grouping them yields

(a⁴ + 2a²b² + b⁴) - 2a²c² - 2b²c² + c⁴ = 0

With a little more effort

(a² + b²)² - 2c²(a² + b²) + c⁴ = 0

And finally

[(a² + b²) - c²]² = 0

Remark

For a quadrilateral with sides a, b, c and d inscribed in a circle there exists a generalization of Heron's formula discovered by Brahmagupta. In this case, the semiperimeter is defined as p = (a + b + c + d)/2. Then the following formula holds

S² = (p - a)(p - b)(p - c)(p - d)

Since any triangle is inscribable in a circle, we may let one side, say d, shrink to 0. This leads to Heron's formula.

Note: the derivative of the right-hand side of Heron's formula - when equated to zero - also leads to the Pythagorean theorem.

References

W. Dunham, Journey through Genius, Penguin Books, 1991

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16S²	= (a + b + c)(- a + b + c)(a - b + c)(a + b - c)
	= 2a²b² + 2a²c² + 2b²c² - (a⁴ + b⁴ + c⁴)