# 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

*S ^{2}* =

*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

16S^{2} | = (a + b + c)(- a + b + c)(a - b + c)(a + b - c) |

= 2a + 2^{2}b^{2}a + 2^{2}c^{2}b - (^{2}c^{2}a + ^{4}b + ^{4}c)^{4} |

For the right triangle, 16*S*^{2} = 4*a ^{2}b^{2}*. So we have

4*a ^{2}b^{2}*= 2

*a*+ 2

^{2}b^{2}*a*+ 2

^{2}c^{2}*b*- (

^{2}c^{2}*a*+

^{4}*b*+

^{4}*c*)

^{4}Taking all terms to the left side and grouping them yields

(*a ^{4}* + 2

*a*+

^{2}b^{2}*b*) - 2

^{4}*a*- 2

^{2}c^{2}*b*+

^{2}c^{2}*c*= 0

^{4}With a little more effort

(*a ^{2}* +

*b*)

^{2}*- 2*

^{2}*c*(

^{2}*a*+

^{2}*b*) +

^{2}*c*= 0

^{4}And finally

[(*a ^{2}* +

*b*) -

^{2}*c*]

^{2}*= 0*

^{2}### 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

*S ^{2}* = (

*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

|Contact| |Front page| |Contents| |Geometry| |Up|

Copyright © 1996-2018 Alexander Bogomolny64638767 |