Problem 10.8. On the face $ABC$ of the tetrahedron $ABCD$, a point $P$ is marked. Points $A_1, B_1, C_1$ are the projections of point $P$ onto the faces $BCD, ACD, ABD$ respectively. It turns out that $PA_1 = PB_1 = PC_1$. Find $\angle BA_1C$, given that $\angle BC_1D = 136^\circ$ and $\angle CB_1D = 109^\circ$.

![](https://cdn.mathpix.com/cropped/2024_05_06_62ad5020cf8c9b2474f9g-5.jpg?height=487&width=505&top_left_y=1000&top_left_x=468)
Answer: 115.

Solution. Consider triangles $D P A_{1}$ and $D P C_{1}$. They are equal by the leg $\left(P C_{1}=P A_{1}\right)$ and the hypotenuse (common $P D$). From this, we get that $D C_{1}=D A_{1}$.

Considering triangles $B P C_{1}$ and $B P A_{1}$, we similarly obtain $B C_{1}=B A_{1}$. Then triangles $B D C_{1}$ and $B A_{1} D$ are equal by three sides, which means $\angle B C_{1} D=\angle B A_{1} D$.

Similarly, we can obtain $\angle C B_{1} D=\angle C A_{1} D$. Now it is clear that the desired angle is calculated as

$$
\angle B A_{1} C=360^{\circ}-\angle B A_{1} D-\angle C A_{1} D=360^{\circ}-\angle B C_{1} D-\angle C B_{1} D=115^{\circ}
$$

Remark. If we draw a sphere with center at point $P$ and radius $P A_{1}$, then it will touch the faces $D A B, D B C, D C A$. Then the equalities of segments, for example, $D C_{1}=D A_{1}=D B_{1}$, will follow from the equality of the segments of tangents to the sphere.