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
[(a2 + b2) - c2]2 = 0
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.
- W. Dunham, Journey through Genius, Penguin Books, 1991
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