In the isosceles triangle $ABC$, $AB = AC$ and $\angle BAD = 30^{\circ}$, where $D$ is an interior point of side $BC$. Furthermore, the triangle $ADE$ is also isosceles, where $E$ is an interior point of side $AC$. What could be the measure of $\angle EDC$?
If we assume that in the isosceles triangle $A D E$, $A D = A E$, then the angle $E D C \varangle$ can be easily determined based on known angle relationships. Most solvers solved the problem this way and found that $E D C \varangle = 15^{\circ}$.

They received 2 points for their solution.

Many people wondered what would happen if the triangle $A D E$ were isosceles in such a way that $E A = E D$ or $D A = D E$. In these latter two cases, the angle $E D C \varangle$ cannot be determined uniquely. In the first case, we get that $E D C \varangle < 60^{\circ}$, and in the second case, $E D C \varangle < 125^{\circ}$.

Those who thought about the three possible triangles and tried to calculate the desired angle received 2 bonus points. It is worth thinking a bit more about the problem beyond the simple solution!