Proof of Heron's Formula

A derivation of Heron's formula for the area of a triangle using the incircle and tangent length relationships.

Keywords: Heron's formula, area, incircle, inradius, tangent lengths, semi-perimeter

Prerequisites: area-proof, incenter-proof · Difficulty: intermediate

Heron's formula gives the area of a triangle from its three side lengths $a$ , $b$ , $c$ alone:

Area = s (s - a) (s - b) (s - c)

where $s = \frac{a + b + c}{2}$ is the semi-perimeter.

Strategy: route through the incircle. Connecting the incenter $I$ to each vertex splits the triangle into three sub-triangles of height $r$ , giving $Area = r s$ . The tangent lengths give $x = s - a$ , $y = s - b$ , $z = s - c$ , and the right triangles at the tangent points complete the formula.

Step 1: The Incircle and Tangent Points

Start with triangle $A B C$ and its incircle centered at $I$ . The incircle touches each side at a tangent point: $X$ on $B C$ , $Y$ on $C A$ , $Z$ on $A B$ .

Step 2: Heron's Formula

Heron's formula gives the area of a triangle directly from its three side lengths $a$ , $b$ , $c$ :

Area = s (s - a) (s - b) (s - c)

where $s = \frac{a + b + c}{2}$ is the semi-perimeter — no angles or heights needed.

Our proof routes through the incircle: first $Area = r s$ from three sub-triangles on the inradius, then the tangent-length identities $x = s - a$ , $y = s - b$ , $z = s - c$ , then a relation $r^{2} s = (s - a) (s - b) (s - c)$ from the right triangles at the tangent points.

Step 3: Area $= r s$

Connect $I$ to each vertex, splitting the triangle into three sub-triangles. Each has a base (a side of the triangle) and height equal to the inradius $r$ :

Area = \frac{1}{2} a r + \frac{1}{2} b r + \frac{1}{2} cr = \frac{r ( a + b + c )}{2} = r s

Step 4: Tangent Lengths

Tangent segments from a point to a circle are equal in length. Let:

From $A$ : $A Y = A Z = x$
From $B$ : $B Z = B X = y$
From $C$ : $C X = C Y = z$

Then $a = y + z$ , $b = x + z$ , $c = x + y$ .

Adding all three: $a + b + c = 2 (x + y + z) = 2 s$ , so $x + y + z = s$ .

Step 5: Solving for $x$ , $y$ , $z$

From $x + y + z = s$ and $a = y + z$ :

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

These tangent lengths are the key to Heron's formula.

Step 6: Derive Heron's Formula

From the right triangles at each tangent point: $tan (A /2) = r / x$ , $tan (B /2) = r / y$ , $tan (C /2) = r / z$ .

Since $A /2 + B /2 + C /2 = 90°$ , we get $tan (A /2 + B /2) = cot (C /2) = z / r$ . Expanding the left side and simplifying yields $r^{2} (x + y + z) = x y z$ , so:

r^{2} = \frac{x y z}{x + y + z} = \frac{( s - a ) ( s - b ) ( s - c )}{s}

Therefore:

Area^{2} = r^{2} s^{2} = s (s - a) (s - b) (s - c)

Area = s (s - a) (s - b) (s - c)

Step 7: Heron's Formula

We have derived Heron's formula:

Area = s (s - a) (s - b) (s - c)

This remarkable formula gives the area of any triangle knowing only its three side lengths — no angles or heights needed!

Heron's formula gives the area of a triangle from its side lengths alone:

Area = s (s - a) (s - b) (s - c)

The key insight is that $Area = r s$ (from the incircle decomposition), combined with the tangent length relationships $x = s - a$ , $y = s - b$ , $z = s - c$ . ∎