10.7. In triangle $A B C$, the median $A M$ and the angle bisector $B N$ intersect at point $P$. Determine the measures of the angles of triangle $A B C$, given that lines $M N$ and $B C$ are perpendicular to each other, and $B P: A N=3: 2$.
Solution: Let $N Q \| A M, Q \in B C$. From the hypothesis we have $B P=3 x, A N=2 x$. Let $P N=y$. In triangle $B C N$ the segment $M N$ is a median and an altitude, so the triangle is isosceles, and $B N=C N=3 x+y$. Since $P M \| N Q$, by Thales' theorem in

![](https://cdn.mathpix.com/cropped/2024_06_06_92c9d954472f21107076g-1.jpg?height=442&width=669&top_left_y=2159&top_left_x=1259)
triangle $B N Q$, we have $\frac{B M}{M Q}=\frac{3 x}{y}=\frac{C M}{M Q}$. We can write $\frac{C Q}{M Q}=\frac{C M-M Q}{M Q}=\frac{C M}{M Q}-1=\frac{3 x}{y}-1$.

Since $N Q \| A M$, by Thales' theorem in triangle $A C M$, we have $\frac{C Q}{M Q}=\frac{C N}{A N}=\frac{3 x+y}{2 x}=\frac{3}{2}+\frac{1}{2} \cdot \frac{y}{x}$. Therefore $3 \cdot \frac{x}{y}-1=\frac{3}{2}+\frac{1}{2} \cdot \frac{y}{x} . \quad$ Let $\frac{x}{y}=t, \quad t>0$. We obtain the equation $3 t-1=\frac{3}{2}+\frac{1}{2 t} \Leftrightarrow 6 t^{2}-5 t-1=0 \Leftrightarrow\left[\begin{array}{l}t=1, \\ t=-\frac{1}{6}\end{array} \Rightarrow t=1 \Rightarrow y=x\right.$.

We observe that $B N=C N=4 x$. Applying the angle bisector theorem in triangle $A B C$ we get $\frac{A B}{B C}=\frac{A B}{2 B M}=\frac{A N}{C N}=\frac{2 x}{4 x}=\frac{1}{2} \Rightarrow A B=B M$. By the congruence criterion for triangles (SAS), it follows that $\triangle A B N \equiv \triangle M B N$. Triangle $M B N$, being right-angled, results in $m(\angle B A C)=90^{\circ}$. Since $A B=\frac{1}{2} \cdot B C$, it follows that $m(\angle A C B)=30^{\circ}$, and $m(\angle A B C)=60^{\circ}$.

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\text { Answer: } m(\angle B A C)=90^{\circ}, m(\angle A C B)=30^{\circ}, m(\angle A B C)=60^{\circ} \text {. }
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