How to find the base of an isosceles triangle with the two equal sides and their medians known

Question

I know, that the two equal sides are equal to $4\sqrt{10}$ and the medians to each of these sides are equal to $6\sqrt{3}$. I feel like, there's some information missing. How do I solve this problem?

Answer 1

You can use a corollary of the Apollonius Theorem to show that the length of the median onto side $a$ is $$m_a = \frac12 \sqrt{2b^2+2c^2-a^2}$$ and since $a=b=4\sqrt{10}$ and $m_a = 6\sqrt{3}$ you are done.

Answer 2

Let the triangle be $ABC$ with $AB=AC$ and $M$ the midpoint of $AC$.

Use the cosine formula on each of $AMB,BMC$ using the angle at $M$ in each case. [Note that the two cosines have equal magnitude and opposite signs.]