# Heron's formula: area of triangle

## Statement

For any triangle with sides $a$, $b$ and $c$, the area is given by:

$$A = \sqrt{s(s - a)(s - b)(s - c)} \text{, where } s = \frac{a + b + c}{2}$$

## Proof

Construct triangle with sides $a$, $b$ and $c$. Then draw altitude $h$ on side $b$, dividing $b$ in $b_1$ and $b_2$. See the image below.

The area of this triangle is given by:

$$A = \tfrac{1}{2} \cdot b \cdot h$$

Using the Pythagorean theorem, find that:

$$a^2 = h^2 + b_1^2 \implies h^2 = a^2 - b_1^2$$

$$c^2 = h^2 + b_2^2 \implies h^2 = c^2 - b_2^2$$

Since both equations equal $h^2$, set them equal to eachother and then isolate the $b$-terms:

$$a^2 - b_1^2 = c^2 - b_2^2$$

Substitute $b = b_1 + b_2 \implies b_1 = b - b_2$ and expand the brackets:

$$a^2 - (b - b_2)^2 = c^2 - b_2^2$$

$$a^2 - (b^2 - 2b \cdot b_2 + b_2^2) = c^2 - b_2^2$$

$$a^2 - b^2 + 2b \cdot b_2 - b_2^2 = c^2 - b_2^2$$

$$a^2 - b^2 + 2b \cdot b_2 = c^2$$

Solve for $b_2$.

$$2b \cdot b_2 = b^2 + c^2 - a^2$$

$$b_2 = \frac{b^2 + c^2 - a^2}{2b}$$

Use the equation $h^2 = c^2 - b_2^2$ once again and substitute $b_2$.

$$h = \sqrt{c^2 - \left(\frac{b^2 + c^2 - a^2}{2b}\right)^2}$$

When substituting $h$ into the area formula, you get:

$$A = \tfrac{1}{2} \cdot b \cdot \sqrt{c^2 - \left(\frac{b^2 + c^2 - a^2}{2b}\right)^2}$$

## Simplifying the formula

Bring the square to the numerator and the denominator and make a common denominator.

$$A = \tfrac{1}{2}b \cdot \sqrt{\frac{4b^2c^2}{4b^2} - \frac{(b^2 + c^2 - a^2)^2}{4b^2}}$$

$$A = \tfrac{1}{2}b \cdot \sqrt{\frac{4b^2c^2 - (b^2 + c^2 - a^2)^2}{4b^2}}$$

Bring the $\frac{1}{2}b$ factor in the square root and simplify.

$$A = \sqrt{\tfrac{1}{4}b^2} \cdot \sqrt{\frac{4b^2c^2 - (b^2 + c^2 - a^2)^2}{4b^2}}$$

$$A = \sqrt{\tfrac{1}{4}b^2 \cdot \frac{4b^2c^2 - (b^2 + c^2 - a^2)^2}{4b^2}}$$

$$A = \sqrt{\frac{4b^2c^2 - (b^2 + c^2 - a^2)^2}{16}}$$

Write $4b^2c^2$ as a square and factor using the difference of squares.

$$A = \sqrt{\frac{(2bc)^2 - (b^2 + c^2 - a^2)^2}{16}}$$

$$A = \sqrt{\frac{(2bc + b^2 + c^2 - a^2)(2bc - b^2 - c^2 + a^2)}{16}}$$

Rearrange the terms and factor out perfect squares.

$$A = \sqrt{\frac{\bigg(b^2 + 2bc + c^2 - a^2\bigg)\bigg(a^2 - (b^2 - 2bc + c^2)\bigg)}{16}}$$

$$A = \sqrt{\frac{\bigg((b + c)^2 - a^2\bigg)\bigg(a^2 - (b - c)^2\bigg)}{16}}$$

Factor using the difference of squares once again.

$$A = \sqrt{\frac{\bigg((b + c - a)(b + c + a)\bigg)\bigg((a + b - c)(a - b + c)\bigg)}{16}}$$

$$A = \sqrt{\frac{(b + c - a)(b + c + a)(a + b - c)(a - b + c)}{16}}$$

Distribute the $16 = 2^4$ to each factor.

$$A = \sqrt{\frac{b + c - a}{2} \cdot \frac{b + c + a}{2} \cdot \frac{a + b - c}{2} \cdot \frac{a - b + c}{2}}$$

Let $s = \dfrac{a + b + c}{2}$.

Note that:

$$\frac{b + c - a}{2} = \frac{b + c + a - 2a}{2} = \frac{b + c + a}{2} - a = s - a$$

$$\frac{a + b - c}{2} = \frac{a + b + c - 2c}{2} = \frac{a + b + c}{2} - c = s - c$$

$$\frac{a - b + c}{2} = \frac{a + b + c - 2b}{2} = \frac{a + b + c}{2} - b = s - b$$

When substituting these, you get:

$$A = \sqrt{(s - a) \cdot s \cdot (s - c) \cdot (s - b)}$$

Finally, rearrange these factors to get:

$$A = \sqrt{s(s - a)(s - b)(s - c)}$$