7. Given a convex quadrilateral $A B C D$. Let $P$ be the center of the circle inscribed in triangle $A B D$, and $Q$ be the center of the circle inscribed in triangle $C B D$. Ray $B P$ intersects side $D A$ at point $M$, and ray $D Q$ intersects side $B C$ at point $N$. It turns out that $A M=\frac{8}{3}, D M=\frac{4}{3}$ and $B N=\frac{6}{5}, C N=\frac{9}{5}$.

a) Find the ratio $A B: C D$.

b) Suppose it is additionally known that the circles mentioned in the problem are tangent to each other. Find the lengths of sides $A B$ and $C D$.
Answer: a) $A B: C D=4: 3$; b) $A B=4, C D=3$.

Solution. a) Since the bisector of a triangle divides its side proportionally to the other two sides, $A B: B D=A M: M D=2: 1$ and $B D: D C=B N: N C=2: 3$. Therefore, $A B: C D=2 \cdot \frac{2}{3}=4: 3$.

b) Let the points of tangency of the incircle of triangle $A B D$ with its sides $A B$, $A D$, $B D$ be $P$, $F$, $K$ respectively; the points of tangency of the incircle of triangle $B C D$ with its sides $B C$, $C D$, $B D$ be $Q$, $E$, $K$ respectively (the point of tangency with side $B D$ is common by condition).

Let $B K=x, K D=y$. Using the equality of the segments of the tangent drawn from one point to a circle, we obtain the relations

$B Q=B P=B K=x, D F=D E=D K=y, A F=A D-D F=4-y, A P=A F=4-y$, $C Q=B C-B Q=3-x, C E=C Q=3-x, A B=A P+P B=4+x-y, C D=3-x+y$.

In part a), it was obtained that $A B: C D=4: 3$, from which $\frac{4+x-y}{3-x+y}=\frac{4}{3}, x=y$. Then $A B=4$, $C D=3$.

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