# How to prove: $S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$

If $$m_a, m_b, m_c$$ are the medians of a triangle and let $$m=\frac{m_a+ m_b+ m_c}{2}$$ then Area $S$ of triangle is given by $$S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$$ This looks very similar to Heron's formula. How to prove this formula?

-

## 2 Answers

$$S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$$ $$=\frac{4}{3}\sqrt{ \frac{(m_a+m_b+m_c)}{2}\frac{(-m_a+m_b+m_c)}{2}\frac{(m_a-m_b+m_c)}{2}\frac{(m_a+m_b-m_c)}{2}}$$ $$=\frac{1}{3}\sqrt{(m_a+m_b+m_c)(-m_a+m_b+m_c)(m_a-m_b+m_c)(m_a+m_b-m_c)}$$ $$=\frac{1}{3}\sqrt{[(m_a+m_b)^2-m_c^2] [m_c^2-(m_b-m_a)^2] }$$ $$=\frac{1}{3}\sqrt{-[(m_a+m_b)(m_b-m_a)]^2+m_c^2(m_a^2+m_b^2)+m_c^2(m_b^2-m_a^2)-m_c^4}$$ $$=\frac{1}{3}\sqrt{-[m_b^2-m_a^2]^2+m_c^2(m_a+m_b)^2+m_c^2(m_b-m_a)^2-m_c^2}$$ $$=\frac{1}{3}\sqrt{2(m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2)-(m_a^4+m_b^4+m_c^4)}$$ Now replacing $m_a$, $m_b$, $m_c$ respectively with $\frac{1}{2}\sqrt{2c^2+2b^2-a^2}$, $\frac{1}{2}\sqrt{2a^2+2c^2-b^2}$, $\frac{1}{2}\sqrt{2a^2+2b^2-c^2}$

You'll arrive at the Heron's Formula:

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

## EDIT

Proof for $m_a = \frac{1}{2}\sqrt{2c^2+2b^2-a^2}$: Assumptions: $BC = a$, $AC = b$, $AD = DE = m_a$ and $AB = c$

From cosine theorem, we have:

$$a^2 = b^2 + c^2 - 2bc*cos A$$ Now, in triangle ABE(since ACD is congruent to EBD);

$$4m_a^2 = AE^2 = b^2 + c^2 - 2bc*cos(\pi - A)$$ $$a^2 + m_a^2 = 2(b^2 + c^2) - 2[bc*cos A + bc*cos(\pi - A)]$$

or,

$$2m_a^2 = 2b^2 + 2c^2 - a^2$$

Hence,

$$m_a = \frac{1}{2}\sqrt{2b^2 + 2c^2 - a^2}$$

-
Explain why for example $m_a=\frac{1}{2}\sqrt{2c^2+2b^2-a^2}$ – Milingona Ana Jul 10 '12 at 9:33
@MilingonaAna Edited the reply. – hjpotter92 Jul 12 '12 at 8:13

A simple way to prove this is given here without much computation: http://jwilson.coe.uga.edu/emt725/Medians.Triangle/Area.Medians.Tri.html

-