# Heron's Formula: a Proof

The area S of a triangle ABC, with side length a, b, c and semiperimeter s = (a + b + c)/2, is given by S² = s(s - a)(s - b)(s - c).

Heron's formula is named after Hero of Alexandria (1 century AD. The formula is a specialization of Brahmagupta's formula for cyclic quadrilaterals. For, after all, every triangle is a cyclic quadrilateral with two coalesced vertices. I find the proof presented below rather amusing because it exploits the dissection of a triangle induced by the presence of the incircle.

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In ΔABC, the lengths of the segments from vertices to the points of tangency of the incircle are found to be

x = s - a, y = s - b, z = s - c,

so that Heron's formula can be also written as S² = sxyz.

Let r be the inradius of ΔABC. The rearrangement of the six triangles of the dissection as done at the bottom of the applet, shows immediately that S = rs.

Let I be the incenter and denote w = AI. From the diagram in the right portion of the applet,

xyz = r²(x + y + z) = r²s.

It then follows that sxyz = r²s² = S², which completes the proof.

Note: let the angles of the triangle be 2α, 2β, 2γ so that α + β + γ = 90°. The identity xyz = r²(x + y + z) is equivalent to the following trigonometric formula:

cotα + cotβ + cotγ = cotα cotβ cotγ,

where "cot" denotes the standard cotangent function.

Note: Heron's formula is an immediate consequence of that of Brahmagupta which is stated for cyclic quadrilaterals. Letting one of the sides vanish leads to Heron's formula. Curiously, Brahmagupta's formula can be derived from Heron's.

### References

1. R. Vakil, A Mathematical Mosaic, Brendan Kelly Publishing, 2008 (Expanded Edition), pp. 178-179 • 