35. Answer: 12
Let the extensions of $A Q$ and $B P$ meet at the point $R$. Then $\angle P R Q=\angle P A B=$ $\angle Q P R$ so that $Q P=Q R$. Since $Q A=Q P$, the point $Q$ is the midpoint of $A R$. As $A R$ is parallel to $L P$, the triangles $A R B$ and $L P B$ are similar so that $M$ is the midpoint of $P L$. Therefore, $N$ is the centroid of the triangle $P L B$, and $3 M N=B M$.

Let $\angle A B P=\theta$. Thus $\tan \theta=A R / A B=2 A Q / A B=5 / 6$. Then $B L=P B \cos \theta=$ $A B \cos ^{2} \theta$. Also $B M / B L=B Q / B A$ so that $3 M N=B M=\frac{B Q}{A B} A B \cos ^{2} \theta=$ $\cos ^{2} \theta(Q M+3 M N)$. Solving for $M N$, we have $M N=\frac{Q M}{3 \tan ^{2} \theta}=\frac{25}{3 \times(5 / 6)^{2}}=12$.
35. In the diagram below, $P$ is a point on the semi-circle with diameter $A B$. The point $L$ is the foot of the perpendicular from $P$ onto $A B$, and $K$ is the midpoint of $P B$. The tangents to the semicircle at $A$ and at $P$ meet at the point $Q$. It is given that $P L$ intersects $Q B$ at the point $M$, and $K L$ intersects $Q B$ at the point $N$. Suppose $\frac{A Q}{A B}=\frac{5}{12}$, $Q M=25 \mathrm{~cm}$ and $M N=x \mathrm{~cm}$. Find the value of $x$.