Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 7 down vote accepted

$$S=\frac{4}{3}\sqrt{ m(m-m_a)(m-m_b)(m-m_c)}$$ $$ =\frac{4}{3}\sqrt{ \frac{(m_a+m_b+m_c)}{2}\frac{(-m_a+m_b+m_c)}{2}\frac{(m_a-m_b+m_c)}{2}\frac{(m_a+m_b-m_c)}{2}}$$ $$ =\frac{1}{3}\sqrt{(m_a+m_b+m_c)(-m_a+m_b+m_c)(m_a-m_b+m_c)(m_a+m_b-m_c)}$$ $$ =\frac{1}{3}\sqrt{[(m_a+m_b)^2-m_c^2] [m_c^2-(m_b-m_a)^2] } $$ $$ =\frac{1}{3}\sqrt{-[(m_a+m_b)(m_b-m_a)]^2+m_c^2(m_a^2+m_b^2)+m_c^2(m_b^2-m_a^2)-m_c^4}$$ $$ =\frac{1}{3}\sqrt{-[m_b^2-m_a^2]^2+m_c^2(m_a+m_b)^2+m_c^2(m_b-m_a)^2-m_c^2}$$ $$ =\frac{1}{3}\sqrt{2(m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2)-(m_a^4+m_b^4+m_c^4)}$$ 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:

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


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

Median extended

From cosine theorem, we have:

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

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


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


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

share|cite|improve this answer
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:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.