Problem 7.8. Given an isosceles triangle $ABC (AB = BC)$. On the ray $BA$ beyond point $A$, point $E$ is marked, and on side $BC$, point $D$ is marked. It is known that

$$
\angle ADC = \angle AEC = 60^{\circ}, AD = CE = 13.
$$

Find the length of segment $AE$, if $DC = 9$.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4986512fdb0330805a0g-23.jpg?height=404&width=293&top_left_y=93&top_left_x=580)
Answer: 4.

Solution. Mark point $K$ on ray $B C$ such that $B E=B K$. Then $A E=C K$ as well.

![](https://cdn.mathpix.com/cropped/2024_05_06_d4986512fdb0330805a0g-23.jpg?height=400&width=297&top_left_y=644&top_left_x=578)

Notice that triangles $A C E$ and $C A K$ are congruent by two sides ($A E=C K, A C$ - common side) and the angle between them ($\angle C A E=\angle A C K$ - adjacent to the equal base angles of the isosceles triangle). Therefore, $A K=C E=13$ and $\angle A K C=\angle A E C=60^{\circ}$.

In triangle $A D K$, the angles at vertices $D$ and $K$ are $60^{\circ}$, so it is equilateral, and $D K=A K=A D=13$. Therefore, $A E=C K=D K-D C=13-9=4$.

## 8th grade