Given is $\triangle ABC$ with the medians $AD$, $BE$ with $|AD|=|BE|$. The medians intersect in $S$.
a. Use similar triangles to show that $|AS|:|SD|=|BS|:|SE|=2:1$.
b. Prove that $\triangle ABC$ is an isosceles triangle.
a. $\triangle ABC\sim\triangle EDC$, so $|AC|:|CE|=|BC|:|CD|=2:1$ (because $D$ and $E$ are the middle of the sides of the triangle).
$\triangle ABS \sim \triangle DES$, and because $|AB|:|DE|=2:1$, $|AS|:|SD|$ and $|BS|:|SE|$ must also be $2:1$.
b. I see a lot of congruent and similar triangles, and a lot of useful information, but I have no idea where to start and which information I should use for my proof. I know I should end with: $|AC|=|BC|$ or $\angle A=\angle B$. Can someone give me a hint?