5. It is known that in the pyramid $A B C D$ with vertex $D$, the sum $\angle A B D+\angle D B C=\pi$. Find the length of the segment $D L$, where $L$ is the base of the bisector $B L$ of triangle $A B C$, if it is known that

$$
A B=9, B C=6, A C=5, D B=1 .
$$
Answer: 7.

Solution. Let point $M$ lie on the line, outside the segment $B C$, beyond point $B$. From the condition, we have the equality of angles $\angle A B D=\angle D B M$. Then the projection of line $B D$ onto the plane $A B C$ is the bisector $B K$ of angle $A B M$. The bisectors $B K$ and $B L$ are perpendicular, as the bisectors of the internal and external angles. Also, line $B L$ is perpendicular to the height of the pyramid $D H$. Therefore, $B L \perp D B K$ and, consequently, $B L \perp D B$, and triangle $L B D$ is a right triangle with $\angle B=\pi / 2$. Similarly, consider the case when point $M$ is beyond point $C$. From the ratio $C L / L A=B C / B A=2 / 3$, we find that $L C=2, A L=3$. Since $B L^{2}=A B \cdot B C-A L \cdot L C$, then $B L^{2}=48$. By the Pythagorean theorem, we get $D L^{2}=D B^{2}+B L^{2}=49$. Answer: 7.

## Grading Criteria.

Guessed the solution: 0 points.

Found the length of the bisector: 1 point.

Determined that triangle $L B D$ is a right triangle: 4 points.