4. On the side $B C$ of triangle $A B C$, a point $M$ is taken such that $B M: M C=2: 7$. The bisector $B L$ of the given triangle and the segment $A M$ intersect at point $P$ at an angle of $90^{\circ}$.

a) Find the ratio of the area of triangle $A B P$ to the area of quadrilateral $L P M C$.

b) On the segment $M C$, a point $T$ is marked such that $M T: T C=1: 6$. It is additionally known that the lines $L T$ and $B C$ are perpendicular. Find the angle $C B L$.
Answer. a) $11: 70$, b) $\arccos \frac{\sqrt{11}}{2 \sqrt{3}}$.

Solution. a) In triangle $A B M$, segment $B P$ is both a bisector and an altitude, so triangle $A B M$ is isosceles, and $B P$ is also its median. Let $B M=2 x$, then $A B=2 x, M C=7 x$. By the property of the bisector of a triangle, $A L: L C=A B: B C=2 x: 9 x=2: 9$.

Let the area of triangle $A B C$ be $S$. Then $S_{A B P}=\frac{1}{2} S_{A B M}=\frac{1}{2} \cdot \frac{2}{9} S_{A B C}=\frac{1}{9} S$. By the theorem on the ratio of areas of triangles, we have $\frac{S_{A P L}}{S_{A M C}}=\frac{A P}{A M} \cdot \frac{A L}{A C}=\frac{1}{2} \cdot \frac{2}{11}=\frac{1}{11}$, hence $S_{A P L}=\frac{1}{11} S_{A M C}=\frac{1}{11} \cdot \frac{7}{9} S, S_{L P M C}=\frac{10}{11} S_{A M C}=\frac{10}{11} \cdot \frac{7}{9} S=\frac{70}{99} S$. The desired ratio is $\frac{1}{9} S: \frac{70}{99} S=\frac{11}{70}$.

b) Since triangles $A B P$ and $A L P$ share a common height from vertex $A$, we have $B P: P L=S_{A B P}: S_{A L P}=\frac{1}{9}: \frac{7}{9 \cdot 11}=11: 7$. Let $B P=11 y, P L=7 y$.

Let $\angle C B L=\gamma$. Then from triangle $B P M$ we get $\cos \gamma=\frac{11 y}{2 x}$, and from triangle $B F L$ we get $\cos \gamma=\frac{3 x}{18 y}$. Equating these expressions for cosine, we find that $x=y \sqrt{33}$, hence $\cos \gamma=\frac{\sqrt{11}}{2 \sqrt{3}}$.