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 expresses the area of a triangle in terms of its three side lengths $a$, $b$, $c$:

\text{Area} = \sqrt{s⁢(s-a)(s-b)(s-c)}
    

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

Derivation Strategy

1. The incircle touches all three sides. Connect the incenter $I$ to each vertex, dividing the triangle into three sub-triangles.
2. Each sub-triangle has height equal to the inradius $r$, so $\text{Area} = rs$.
3. The tangent lengths from each vertex give $x = s - a$, $y = s - b$, $z = s - c$.
4. Combining with the Pythagorean theorem applied to the right triangles formed at each tangent point, we derive the full formula.

The Proof

Step 1: The Incircle and Tangent Points

Start with triangle $ABC$ and its incircle centered at $I$. The incircle touches each side at a tangent point: $X$ on $BC$, $Y$ on $CA$, $Z$ on $AB$.

Step 2: Heron's Formula

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

\text{Area} = \sqrt{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 $\text{Area} = rs$ 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 $= rs$

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 $r$:

\text{Area} = \frac{1}{2}ar + \frac{1}{2}br + \frac{1}{2}cr
    = \frac{r⁢(a+b+c)}{2} = rs
    

Step 4: Tangent Lengths

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

• From $A$: $AY = AZ = x$
• From $B$: $BZ = BX = y$
• From $C$: $CX = CY = z$

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

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

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

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

x = s - a, \quad y = s - b, \quad 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 $r^2(x+y+z) = xyz$, so:

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

Therefore:

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

\text{Area} = \sqrt{s⁢(s-a)(s-b)(s-c)}
    

Step 7: Heron's Formula

We have derived Heron's formula:

\text{Area} = \sqrt{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!

Conclusion

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

\text{Area} = \sqrt{s⁢(s-a)(s-b)(s-c)}
    

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