8.4. In triangle $A B C$, the bisector $A M$ is perpendicular to the median $B K$. Find the ratios $B P: P K$ and $A P: P M$, where $P$ is the point of intersection of the bisector and the median.
Answer: $\quad B P: P K=1, A P: P M=3: 1$. Solution. Triangles $A B P$ and $A K P$ are equal (side $A P$ is common, and the adjacent angles are equal by condition). Therefore, $A B=A K=K C, B P=P K$. Hence, triangles $A B M$ and $A K M$ are equal (by two sides and the angle $\frac{\angle A}{2}$ between them). Then $S_{A B M}=S_{A K M}=S_{C K M}==\frac{S}{3}$, where $S=S_{A B C}$. Further, $S_{A B P}=\frac{1}{2} S_{A B K}=\frac{1}{4} S$. But $S_{A B M}=\frac{1}{2} A M \cdot B P$ (since $A M \perp B P), \quad S_{A B P}=\frac{1}{2} A P \cdot B P$. From this, $A M=\frac{2 S}{3 \cdot B P}, A P=\frac{2 S}{4 \cdot B P}$, i.e., $\frac{A P}{A M}=\frac{3}{4}$ and therefore $A P$ : $P M=1: 3$.

# 9th grade