In the drawing below, $C$ is the intersection point of $A E$ and $B F$, $A B=B C$, and $C E=C F$. If $\angle C E F=50^{\circ}$, determine the angle $\angle A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_01_20684006194a352111ccg-10.jpg?height=552&width=885&top_left_y=1363&top_left_x=491)
Solution

Since triangle $CEF$ is isosceles, we have $\angle CEF = \angle CFE = 50^{\circ}$ and

$$
\begin{aligned}
\angle FCE & =180^{\circ}-\angle CEF-\angle CFE \\
& =180^{\circ}-50^{\circ}-50^{\circ} \\
& =80^{\circ}
\end{aligned}
$$

We also have $\angle BCA = \angle FCE$, as they are vertically opposite angles. Furthermore, since $ABC$ is isosceles, we have $\angle BAC = \angle BCA = 80^{\circ}$. Finally, we can conclude that

$$
\begin{aligned}
\angle ABC & =180^{\circ}-\angle BAC-\angle BCA \\
& =180^{\circ}-80^{\circ}-80^{\circ} \\
& =20^{\circ}
\end{aligned}
$$