[ Pythagorean Theorem (direct and inverse). Similarity criteria. In triangle $ABC$, a perpendicular line passing through the midpoint of side $AB$ intersects the extension of side $BC$ at point $M$, such that $MC: MB = 1: 5$. A perpendicular line passing through the midpoint of side $BC$ intersects side $AC$ at point $N$, such that $AN: NC = 1: 2$. Find the angles of triangle $ABC$.

#
Continue the perpendicular bisector of side $B C$ until it intersects with line $A B$ and with the line drawn through vertex $C$ parallel to $A B$.

## Solution

Let $P$ and $Q$ be the midpoints of sides $B C$ and $A B$ respectively, and let $K$ be the point of intersection of lines $P N$ and $A B$. Draw a line through point $C$ parallel to $A B$ until it intersects with line $P N$ at point $T$. Denote $M C=x, A Q=a$. Then $C P=P B=2 x, B Q=a$.

From the congruence of triangles $K P B$ and $T P C$, it follows that $C T=K B$, and from the similarity of triangles $K N A$ and $T N C$, we have $C T=N A: N C=1: 2$. Therefore, $K A=A B=2 a$.

From the similarity of triangles $M B Q$ and $K B P$, it follows that $P B: Q B=$ $K B : M B$, or $2 x / a=4 a / 5 x$. Hence, $4 a^{2}=10 x^{2}$. Let $A F$ be the height of triangle $A B C$. Then $A F$ is the midline of triangle $K P B$. Therefore, $B F=F P=x$. Thus,

$A F^{2}=A B^{2}-B F^{2}=4 a^{2}-x^{2}=10 x^{2}-x^{2}=9 x^{2}$, so $A F=3 x$.

Therefore, $\operatorname{tg} \angle C=A F / F C=1, \angle C=45^{\circ}$, and $\operatorname{tg} \angle B=A F / F B=3$. Hence, $\operatorname{tg} \angle A=-\operatorname{tg}(\angle A+\angle B)=-\frac{1+3}{1-3}$

![](https://cdn.mathpix.com/cropped/2024_05_06_7181633d50815d79481eg-35.jpg?height=432&width=1212&top_left_y=1&top_left_x=432)

## Answer

$\angle A=\operatorname{arctg} 2, \angle B=\operatorname{arctg} 3, \angle C=45^{\circ}$.