12.53. In triangle $ABC$, the angle bisectors $AD$ and $BE$ are drawn. Find the measure of angle $C$, given that $AD \cdot BC = BE \cdot AC$ and $AC \neq BC$.
12.53. The quantities $A D \cdot B C \sin A D B$ and $B E \cdot A C \sin A E B$ are equal, as they are equal to twice the area of triangle $A B C$. Therefore, $\sin A D B = \sin A E B$. There are two possible cases.

1. $\angle A D B = \angle A E B$; in this case, points $A, E, D, B$ lie on the same circle, so $\angle E A D = \angle E B D$, i.e., $\angle A = \angle B$, which cannot be true according to the condition.
2. $\angle A D B + \angle A E B = 180^{\circ}$; in this case, $\angle E C D + \angle E O D = 180^{\circ}$, where $O$ is the point of intersection of the angle bisectors. Since $\angle E O D = 90^{\circ} + \angle C / 2$ (Problem 5.3), then $\angle C = 60^{\circ}$.