3. Chords $AB$ and $CD$ of a circle with center $O$ have a length of 5. The extensions of segments $BA$ and $CD$ beyond points $A$ and $D$ respectively intersect at point $P$, and $DP=13$. Line $PO$ intersects segment $AC$ at point $L$. Find the ratio $AL: LC$.
Answer: $A L: L C=13: 18$.

Solution. Drop perpendiculars $O H$ and $O N$ from point $O$ to the chords $C D$ and $A B$ respectively. Since these chords are equal, the distances from the center of the circle to them are also equal, so $O H=O N$. Right triangles $O P N$ and $O P H$ are equal by the leg and hypotenuse ( $O P$ is common), therefore $P N=P H$, $\angle O P N=\angle O P H$ (the latter means that $P O$ is the bisector of angle $B P C$). Since a diameter perpendicular to a chord bisects it, points $H$ and $N$ are the midpoints of $C D$ and $A B$ respectively; hence $A N=D H$ and therefore $A P=D P=13$. Since $P L$ is the bisector of triangle $A P C$, we get that $\frac{A L}{L C}=\frac{P A}{P C}=\frac{P A}{P D+D C}=\frac{13}{18}$.