Triangle with a median

Question

In the triangle $ABC$ the length of a median beginning in A is $\sqrt{10}$. If $|AB|=4$ and $|AC|=6$ what is the length of BC?

I tried using the law of cosines with different parts of the angle near A but without any significant results. Could you help?

Answer 1

Hint: Try Stewart's Theorem. It is

$$dad + man = bmb + cnc$$

where $d$ is the length of the median, and $m+n = a$ are the lengths created by the median to the side of length $a$.

Answer 2

Use Al Kashi's theorem: let $x=\frac 12 BC$ and $I$be the midpoint of $BC$. Applying Al Kashi's theorem in triangles $AIB$ and $AIC$ yields $$\begin{cases} 16=x^2+10-2\sqrt{10}x\\ 36=x^2+10+2\sqrt{10}x \end{cases}\enspace\text{ whence }\; 2\sqrt{10} x=x^2-6=26-x^2$$ so $\;x^2=16$ and eventually $BC=8$.