AL and BM are bisectors in $\triangle ABC$. The second common point of the circles described (circumscribed) around $\triangle ACL$ and $\triangle BCM$ lies on the side AB. Find $\angle ACB$.
I am not sure about the word "described" in the problem and if I am not right, I would be very grateful if you correct me.
Look at the drawing to understand what I have noticed. $$\alpha + \beta = 90^\circ - \frac{\gamma}{2} $$ $$ \angle AFB = 90^\circ + \frac{\gamma}{2}$$
I am not sure that these statements will help with the solution but I just want to show I have tried sth.
Hint:
$$\begin {cases}\angle AEC=\angle ALC=\pi-\alpha-\gamma\\ \angle BEC=\angle BMC=\pi-\beta-\gamma\\ \alpha+\beta=\pi/2-\gamma/2 \end {cases}\\ \implies \pi=\angle AEC+\angle BEC=(\pi-\alpha-\gamma)+(\pi-\beta-\gamma)=\frac32 (\pi-\gamma). $$
Can you take it from here?