Similarity Signs

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 $K$ 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 $T$. Let $M C=x, A Q=a$. Then $C P=P B=2 x, B Q=a$.

From the congruence of triangles $K R B$ 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 $K A : C T = N A : N C = 1 : 2$. 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 $\frac{2 x}{a} = \frac{4 a}{5 x}$. Hence, $4 a^2 = 10 x^2$. Let $A F$ be the altitude of triangle $A B C$. Then $A F$ is the midline of triangle $K R B$. 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, $\operatorname{tg} \angle C = \frac{A F}{F C} = 1, \angle C = 45^{\circ}$, and $\operatorname{tg} \angle B = \frac{A F}{F B} = 3$. Hence, $\operatorname{tg} \angle A = -\operatorname{tg}(\angle A + \angle B) = -\frac{1 + 3}{1 - 3} = 2$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-20.jpg?height=457&width=1196&top_left_y=2221&top_left_x=431)

## Answer

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

## Problem