Find the lengths of the segments that are formed when the angle bisectors intersect the median in a right triangle

Question

The angle bisectors $AL$ and $BQ$ in the right triangle $\triangle ABC$ with $\angle ACB=90^\circ$ and catheti $15$ and $20$ divide the median $CM$ into three segments. Find their lengths.

I am not sure how to approach the problem. The most "obvious" thing that we can do is to find the hypotenuse $c=\sqrt{a^2+b^2}=25$ of the triangle $\triangle ABC$. Therefore, $m_c=\dfrac12 \cdot 25=12.5$. On my sketch $AC=15$ and $BC=20$. What to do next?

Answer 1

Let $CM\cap AL=\{P\}$ and $BQ\cap CM=\{T\}$.

Thus, $$\frac{CP}{PM}=\frac{AC}{AM}$$ and $$\frac{CT}{TM}=\frac{BC}{BM}.$$ Can you end it now?