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