Geometric proof with a isosceles triangle

Question

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.

My try:

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?

Answer 1

Karolis Juodelė has already provided an answer, but it seems that you need more. So, I'll give you a hint.

You've already proved that $$|AS|:|SD|=|BS|:|SE|=2:1,$$ i.e. $$|AS|=2|SD|\quad\text{and}\quad |BS|=2|SE|.$$ Now use that $|AD|=|BE|$, i.e. $|AS|+|SD|=|BS|+|SE|$ to obtain $$3|SD|=3|SE|\quad\Rightarrow\quad |SD|=|SE|\quad \text{and}\quad |AS|=|BS|$$ from which $\triangle{SEA}\cong\triangle{SDB}$ follows.

Answer 2

You know that $\triangle ABS$ is isosceles. Deduce that $\triangle AES \sim \triangle BDS$ and thus $\triangle ABD \sim \triangle ABE$. Therefore $\angle A = \angle B$.