4. On the side $B C$ of triangle $A B C$, a point $M$ is taken such that $B M: M C=3: 7$. 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 $M T: T C=1: 6$. Suppose it is additionally known that the lines $L T$ and $B C$ are perpendicular. Find the angle $C B L$.
Answer. a) $39: 161$, b) $\arccos \sqrt{\frac{13}{15}}$.

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 $B M=3 x$, then $A B=3 x, M C=7 x$. By the property of the bisector of a triangle, $A L: L C=A B: B C=3 x: 10 x=3: 10$.

Let the area of triangle $A B C$ be $S$. Then $S_{A B P}=\frac{1}{2} S_{A B M}=\frac{1}{2} \cdot \frac{3}{10} S_{A B C}=\frac{3}{20} S$. By the theorem on the ratio of areas of triangles, we have $\frac{S_{A P L}}{S_{A M C}}=\frac{A P}{A M} \cdot \frac{A L}{A C}=\frac{1}{2} \cdot \frac{3}{13}=\frac{3}{26}$, hence $S_{A P L}=\frac{3}{26} S_{A M C}=\frac{3}{26} \cdot \frac{7}{10} S$, and $S_{L P M C}=\frac{23}{26} S_{A M C}=\frac{23}{26} \cdot \frac{7}{10} S=\frac{161}{260} S$. The desired ratio is $\frac{3}{20} S: \frac{161}{260} S=\frac{39}{161}$.

b) Since triangles $A B P$ and $A L P$ share a common height from vertex $A$, we have $B P: P L=S_{A B P}: S_{A L P}=\frac{3}{20}: \frac{21}{260}=13: 7$. Let $B P=13 y, P L=7 y$.

Let $\angle C B L=\gamma$. Then from triangle $B P M$ we get $\cos \gamma=\frac{13 y}{3 x}$, and from triangle $B T L$ - $\cos \gamma=\frac{4 x}{20 y}$. Equating these expressions for cosine, we find that $x=y \sqrt{\frac{65}{3}}$, hence $\cos \gamma=\sqrt{\frac{13}{15}}$.