5. A circle with center $O$, inscribed in triangle $P Q R$, touches its sides $P Q, Q R$ and $R P$ at points $C, A$ and $B$ respectively. Lines $B O$ and $C O$ intersect sides $P Q$ and $P R$ at points $K$ and $L$ respectively. Find the ratio $Q A: A R$, if $K Q=3, Q R=16, L R=1$.
Answer: $9: 7$.

Solution. Triangles $P B K$ and $P C L$ are equal ($P B=P C$ as tangents drawn from the same point to a circle; $\angle P$ is common; angles at vertices $B$ and $C$ are right angles). Therefore, $P K=P L$, and since $P C=P B$, then $C K=B L$. Let $C K=B L=x$. Then $Q A=Q C=$ $Q K+K C=3+x, R A=R B=R L+L B=1+x$. Since $Q R=16$, we get that $(3+x)+(1+x)=16$, from which $x=6$. Thus, $A Q: A R=9: 7$.