In the figure, the triangle $\triangle A B C$ is isosceles with base $B C$ and the angle $\angle B A C$ measures $30^{\circ}$. The triangle $\triangle B C D$ is isosceles with base $B D$. Determine the measure of the angle $\angle D C A$.

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-36.jpg?height=512&width=386&top_left_y=500&top_left_x=719)
Solution

Since $\triangle A B C$ is isosceles, we have $\angle A B C=\angle A C B=\alpha$. Since $\triangle B C D$ is isosceles, we also have $\angle B D C=\angle D B C=\alpha$. If $\angle D C A=\beta$, by the Exterior Angle Theorem, we have $\alpha=30^{\circ}+\beta$. Analyzing the sum of the angles of $\triangle B C D, \alpha+\alpha+\alpha-\beta=180^{\circ}$. Thus, from the two equations we have $2 \beta+90^{\circ}=180^{\circ}$, that is, $\beta=45^{\circ}$. Therefore, the measure of the angle $\angle D C A$ is $45^{\circ}$.