Pythagoras Theorem in Triangles

Question

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.

Thanks.

Answer 1

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:$$AC^2 =\underbrace{AM^2 - BM^2}_{=AB^2} + \underbrace{4BM^2}_{=BC^2} $$ $$\Longrightarrow AC^2 = AM^2 + 3BM^2$$

Answer 2

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*}