Prove Heron's formula

Problem

Prove Heron’s formula $S = \sqrt{s(s-a)(s-b)(s-c)}$, with $s=(a+b+c)/2$, starting from the formulae seen (the cosine rule and the trigonometric area).

Solution

Start from the area $S = \tfrac12 bc\sin\widehat{A}$, whence $S^2 = \tfrac14 b^2c^2\sin^2\widehat{A} = \tfrac14 b^2c^2\left(1-\cos^2\widehat{A}\right)$.

From the cosine rule we obtain $\cos\widehat{A} = \dfrac{b^2+c^2-a^2}{2bc}$, hence: $S^2 = \tfrac14 b^2c^2\left[1 - \left(\frac{b^2+c^2-a^2}{2bc}\right)^2\right] = \tfrac{1}{16}\left[(2bc)^2 - (b^2+c^2-a^2)^2\right].$ The term in brackets is a difference of two squares: $(2bc)^2 - (b^2+c^2-a^2)^2 = \bigl[2bc+(b^2+c^2-a^2)\bigr]\bigl[2bc-(b^2+c^2-a^2)\bigr].$ The first factor is $(b+c)^2 - a^2 = (b+c+a)(b+c-a)$; the second is $a^2-(b-c)^2 = (a+b-c)(a-b+c)$. Introducing the semiperimeter $s=(a+b+c)/2$ we have $b+c+a=2s$, $b+c-a=2(s-a)$, $a+b-c=2(s-c)$, $a-b+c=2(s-b)$. Therefore: $S^2 = \tfrac{1}{16}\cdot 2s\cdot 2(s-a)\cdot 2(s-c)\cdot 2(s-b) = s(s-a)(s-b)(s-c),$ whence $S = \boxed{\sqrt{s(s-a)(s-b)(s-c)}}$. $\blacksquare$

Links

Topics: Triangle trigonometry
Concepts: Triangle area · Heron’s formula · Cosine rule
Skills: Proving
Exercise type: Proof