In the drawing below, points $E$ and $F$ belong to the sides $A B$ and $B D$ of triangle $\triangle A B D$ in such a way that $A E=A C$ and $C D=F D$. If $\angle A B D=60^{\circ}$, determine the measure of the angle $\angle E C F$.

![](https://cdn.mathpix.com/cropped/2024_05_01_360f8ce7ec440aed2c7ag-06.jpg?height=417&width=499&top_left_y=591&top_left_x=687)
Solution

Let $2 \alpha=\angle E A C$ and $2 \beta=\angle F D C$. Since triangles $\triangle E A C$ and $\triangle F D C$ are isosceles, it follows that $\angle A C E=\angle A E C=90^{\circ}-\alpha$ and $\angle D C F=\angle C F D=90^{\circ}-\beta$. Consequently, $\angle E C F=\alpha+\beta$. Now analyzing the sum of the angles of triangle $\triangle A B D$, we have $60^{\circ}+2 \alpha+2 \beta=180^{\circ}$, that is, $60^{\circ}=\alpha+\beta$. Since we already know that $\angle E C F=\alpha+\beta$, then $\angle E C F=60^{\circ}$.