Let $A B C$ be a triangle such that $\angle A C B=50^{\circ}$. The altitude from vertex $A$ and the angle bisector of $\angle A B C$ intersect at $P$, with $P$ inside the triangle $A B C$ and $\angle A P B=105^{\circ}$. Find the measures of the angles $\angle B A C$ and $\angle A B C$.

![](https://cdn.mathpix.com/cropped/2024_05_01_09dbad5281c800d666d6g-29.jpg?height=505&width=1060&top_left_y=1878&top_left_x=570)
Solution

By the Exterior Angle Theorem,

$$
\begin{aligned}
\angle B P A & =105^{\circ} \\
\angle P B T+\angle B T P & =105^{\circ} \\
\angle P B T+90^{\circ} & =105^{\circ} \\
\angle P B T & =15^{\circ}
\end{aligned}
$$

Therefore, since $B P$ is the bisector of $\angle A B C$, it follows that

$$
\angle A B C=2 \cdot \angle P B T=30^{\circ}
$$

Thus, $\angle B A T=180^{\circ}-\angle A B C-\angle A T B=60^{\circ}$. Finally, since $\angle C A T=180^{\circ}-\angle A T C- \angle A C T=40^{\circ}$, it follows that $\angle B A C=60^{\circ}+40^{\circ}=100^{\circ}$. That is, the angles of the triangle are $100^{\circ}, 30^{\circ}$ and $50^{\circ}$.