Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

The question is:

In a triangle $\triangle ABC$, angle $\angle B = 90^\circ$ and $M$ is the mid-point of $BC$. Prove that $|AC|^2 = |AM|^2 + 3|BM|^2$.


Please help me. I have tried various ways but can't figure out a way.


share|cite|improve this question
I should try this approach on my taxes: Dear IRS: I looked at your forms. I tried to fill them out but I couldn't figure out a way. Help. Thanks. – rschwieb Aug 8 '13 at 13:08
up vote 1 down vote accepted

Use Pythagoras' Theorem twice and substitute.

You know from Pythagoras that $AC^2 = AB^2 + BC^2$.

Since $M$ is the mid-point of [BC], $$BC = 2BM, \Longrightarrow BC^2 = 4BM^2$$ Also using Pythagoras on triangle $ABM$ that $$AM^2 = AB^2 + BM^2$$ ie $$AB^2 = AM^2 - BM^2$$

Sub everything in the first equation to get:$$AC^2 =\underbrace{AM^2 - BM^2}_{=AB^2} + \underbrace{4BM^2}_{=BC^2} $$ $$\Longrightarrow AC^2 = AM^2 + 3BM^2$$

share|cite|improve this answer
That was really helpful of you. – Gaurang Tandon Aug 8 '13 at 13:50

Take a look at the triangle $\triangle ABM$ and get $|AB|^2$ from $|AM|^2$ and $|BM|^2$. Then use $|AB|^2$ and $|AC|^2 = (2|BM|)^2$ to get $|AC|^2$.

Edit: Here it is:

Note that $|AB|^2 + |BM|^2 = |AM|^2$, so $|AB|^2 = |AM|^2 - |BM|^2$.

Now, we see that

\begin{align*} |AC|^2 &= |AB|^2 + |BC|^2 = (|AM|^2 - |BM|^2) + (2|BM|)^2 \\ &= |AM|^2 - |BM|^2 + 4|BM|^2 = |AM|^2 + 3|BM|^2. \end{align*}

share|cite|improve this answer
So, I got $|BM|^2 = |AM|^2 - |AB|^2$ .And the final answer which I get is $|AC|^2 = 2|AM|^2 - |AB|^2$,which is wrong... – Gaurang Tandon Aug 8 '13 at 13:16
I said to get $|AB|^2$ from the other two, not $|BM|^2$. Here, I've edited my answer. I hope it helps you. – Vedran Šego Aug 8 '13 at 13:29
Thanks for the help :) – Gaurang Tandon Aug 8 '13 at 13:51

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.