5. In triangle $A B C$, angle $A$ is $75^{\circ}$, and angle $C$ is $60^{\circ}$. On the extension of side $A C$ beyond point $C$, segment $C D$ is laid off, equal to half of side $A C$. Find angle $B D C$.
Answer: $\angle B D C=45^{\circ}$.

Solution. In this triangle, angle $A B C$ is obviously equal to $180^{\circ}-75^{\circ}-60^{\circ}=45^{\circ}$. Draw the altitude $A H$ and note that triangle $A H B$ is isosceles, since $\angle B A H=90^{\circ}-45^{\circ}=\angle A B H$ (Fig. 1). Therefore, $A H=B H$. In the right triangle $A H C$, angle $H A C$ is equal to $75^{\circ}-45^{\circ}=30^{\circ}$, so the leg $H C$ is half the hypotenuse $A C$, which means it is equal to $C D$. Thus, triangle $H C D$ is isosceles, and $\angle H C D=120^{\circ}$, so angles $C H D$ and $C D H$ are equal to $\frac{1}{2}\left(180^{\circ}-120^{\circ}\right)=30^{\circ}$. Since $\angle H A D=\angle A D H=30^{\circ}$, triangle $A H D$ is also isosceles and $B H=A H=D H$, meaning triangle $B H D$ is isosceles as well. By the property of the exterior angle, $\angle H B D=\angle H D B=\frac{1}{2} \angle C H D=15^{\circ}$. Therefore, $\angle B D C=\angle B D H+\angle H D C=45^{\circ}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_2af23d428448f4562f21g-2.jpg?height=507&width=640&top_left_y=1481&top_left_x=772)

Fig. 1

Criteria. Only the answer - 0 points. Isosceles property of triangle $A H B$ - 1 point, triangle $A H D$ - 4 points, triangle $B H D$ - 5 points. Full solution - 7 points.