In the diagram, $A B C D$ is a trapezoid with $B C$ parallel to $A D$ and $A B=C D$. Point $E$ is on $A D$ so that $B E$ is perpendicular to $A D$ and point $F$ is the point of intersection of $A C$ with $B E$. If $A F=F B$ and $\angle A F E=50^{\circ}$, what is the measure of $\angle A D C$ ?

![](https://cdn.mathpix.com/cropped/2024_04_30_790b6557ee0304df033eg-1.jpg?height=396&width=689&top_left_y=1672&top_left_x=772)
In $\triangle A E F$, we have $\angle F A E=180^{\circ}-\angle A E F-\angle A F E=180^{\circ}-90^{\circ}-50^{\circ}=40^{\circ}$.

Since $B F E$ is a straight line, then $\angle A F B=180^{\circ}-\angle A F E=180^{\circ}-50^{\circ}=130^{\circ}$.

Now $\triangle A F B$ is isosceles with $A F=F B$.

Thus, $\angle F A B=\angle F B A=\frac{1}{2}\left(180^{\circ}-\angle A F B\right)=\frac{1}{2}\left(180^{\circ}-130^{\circ}\right)=25^{\circ}$.

Thus, $\angle B A E=\angle F A B+\angle F A E=25^{\circ}+40^{\circ}=65^{\circ}$.

Finally, $\angle A D C=\angle B A E$. (This is because a perpendicular dropped from $C$ to a point $P$ on $A D$ would create $\triangle C P D$ which was congruent to $\triangle B E A$ - these triangles are each rightangled, have equal hypotenuses $B A=C D$ and have equal heights $B E=C P$, since $B C$ and $A D$ are parallel.)

Therefore, $\angle A D C=65^{\circ}$.

ANSWER: $65^{\circ}$