4. (30 points) On the sides $B C$ and $A C$ of the isosceles triangle $A B C (A B = A C)$, points $D$ and $E$ were found respectively such that $A E = A D, \angle E D C = 18^{\circ}$. Find the measure of the angle $\angle B A D$.
Answer: $36^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_4bbf958d7361888899fbg-2.jpg?height=548&width=668&top_left_y=1348&top_left_x=685)

Solution. Denote the angles as indicated in the figure. The angle $A D C$ is an exterior angle for triangle $A D B$; hence, $\beta+18^{\circ}=\alpha+x$. Angles $A B C$ and $A C B$ are the base angles of isosceles triangle $A B C$; hence, $\alpha=\frac{180^{\circ}-(x+y)}{2}$. Angles $A D C$ and $A E D$ are the base angles of isosceles triangle $A D E$; hence, $\beta=\frac{180^{\circ}-y}{2}$. Substitute these expressions into the relation for angle $A D C$:

$$
\frac{180^{\circ}-y}{2}+18^{\circ}=\frac{180^{\circ}-(x+y)}{2}+x
$$

Simplifying the expression, we get $18^{\circ}=\frac{x}{2}$, which means $x=36^{\circ}$.