# Problem 5. (based on materials from the Ural Tournaments)

In triangle $ABC$, angle $B$ is $80^{\circ}$. On side $BC$, point $D$ is marked such that $AB = AD = CD$. On side $AB$, point $F$ is marked such that $AF = BD$. On segment $AC$, point $E$ is marked such that $AB = AE$. Find angle $AEF$.

![](https://cdn.mathpix.com/cropped/2024_05_06_19300bfa006950daee48g-3.jpg?height=1431&width=1697&top_left_y=344&top_left_x=192)
Answer: $20^{\circ}$. Solution:

Since $A B=A D$ we have $\angle A D B=\angle A B D=80^{\circ}$, and since $A D=D C$, we have $\angle D A C=\angle D C A=40^{\circ}$. Therefore, $\angle B A E=60^{\circ}$, hence triangle $A B E-$ is equilateral.

Then denoting by $X$ the intersection point of $A D$ and $B E$ we have $\angle B X D=\angle B A X+\angle A B X=80^{\circ}=\angle B D X$, that is, $B X=B D$.

From this it follows that triangles $A B X$ and $E A F$ are equal by two sides and the angle between them.

Thus, $\angle A E F=\angle B A X=20^{\circ}$.

Criteria: Correct solution: 15 points (+)

Found and proved the equilateral triangle: no less than 6 points (〒)

#