[ Mutual relations between sides and angles of triangles (other).]

In triangle $ABC$, the bisectors $AD$ and $BE$ are drawn. Find the measure of angle $C$, given that $AD \cdot BC = BE \cdot AC$ and $AC \neq BC$.
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 $\underline{5.3}$ ), then $\angle C=60^{\circ}$.