In the following figure, $ABC$ is an isosceles triangle with $BA = BC$. Point $D$ is inside it such that $\angle ABD = 13^\circ$, $\angle ADB = 150^\circ$, and $\angle ACD = 30^\circ$. Additionally, $ADE$ is an equilateral triangle. Determine the value of the angle $\angle DBC$.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-38.jpg?height=589&width=553&top_left_y=1614&top_left_x=633)
Solution

Draw the segments indicated in the following figure.

![](https://cdn.mathpix.com/cropped/2024_05_01_4fd1b602786ad9981479g-39.jpg?height=664&width=555&top_left_y=707&top_left_x=825)

Since $\triangle A D E$ is equilateral, it follows that $\angle A D E=60^{\circ}$. Therefore, $\angle B D E=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$. Moreover, from $A D=D E, B D=B D$ and $\angle A D B=\angle B D E$, it follows by the $L A L$ congruence case that $\triangle A B D \equiv \triangle B D E$. Thus, $A B=B E$ and $\angle A B E=2 \cdot \angle A B D=26^{\circ}$. Now consider the circle $k$ that passes through $A, D$ and $C$. The angle $\angle A C D$ is inscribed in the arc corresponding to the chord $A D$, and thus the center of this circle produces a central angle of $60^{\circ}$ with respect to this chord. That is, $E$ is the center of $k$. Therefore, $A E=E D=E C$. Finally, from $B A=B C$, since the sides of triangles $A B E$ and $B C E$ are congruent, the $L L L$ congruence case produces $\triangle A B E \equiv \triangle B C E$ and this gives $\angle A B E=\angle E B C$. Therefore,

$$
\angle D B C=\angle D B E+\angle E B C=13^{\circ}+26^{\circ}=39^{\circ} \text {. }
$$