In an isosceles triangle $A B C(A B=B C)$, the medians $A M$ and $C N$ intersect at point $D$ at a right angle. Find all the angles of triangle $A B C$ and the area of quadrilateral $N B M D$, if the base $A C=1$.

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The medians of a triangle intersect at one point and are divided by it in the ratio 2:1, counting from the vertex.

## Solution

Since the triangle is isosceles, its medians drawn to the lateral sides are equal, i.e., $A M = C N$.

The medians of a triangle intersect at one point and are divided by it in the ratio 2:1, counting from the vertex, so $C D = \frac{2}{3} \cdot C N = \frac{2}{3} \cdot A M = A D$.

From the isosceles right triangle $A D C$, we find that $C D = A C \cos 45^{\circ} = \frac{\sqrt{2}}{2}$. Then

$$
D M = \frac{1}{2} \cdot A D = \frac{1}{2} \cdot C D = \frac{\sqrt{2}}{4}
$$

From the right triangle $M D C$, we find that

$$
C M = \sqrt{C D^{2} + D M^{2}} = \sqrt{\left(\frac{\sqrt{2}}{2}\right)^{2} + \left(\frac{\sqrt{2}}{4}\right)^{2}} = \sqrt{\frac{5}{8}}
$$

Thus, $B C = 2 \cdot C M = 2 \sqrt{\frac{5}{8}} = \sqrt{\frac{5}{2}}$.

Let $B H$ be the height of triangle $A B C$. Then $H$ is the midpoint of $A C$. From the right triangle $B H C$, we find that

$$
B H = \sqrt{B C^{2} - C H^{2}} = \sqrt{\left(\sqrt{\frac{5}{2}}\right)^{2} - \left(\frac{1}{2}\right)^{2}} = \frac{3}{2}
$$

Therefore,

$$
\operatorname{tg} \angle B A C = \operatorname{tg} \angle B C A = \frac{B H}{C H} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3
$$

Since $M N$ is the midline of triangle $A B C$, then $M N = \frac{1}{2} \cdot A C = \frac{1}{2}$ and $M N \parallel A C$. Therefore, $M N \perp B H$.

The diagonals $M N = \frac{1}{2}$ and $B D = \frac{2}{3} \cdot B H = 1$ of quadrilateral $N B M D$ are perpendicular, so

$$
S_{\mathrm{NBMD}} = \frac{1}{2} \cdot N M \cdot B D = \frac{1}{2} \cdot \frac{1}{2} \cdot 1 = \frac{1}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_d1936174aeda27dcc10dg-40.jpg?height=736&width=534&top_left_y=-5&top_left_x=759)

## Answer

$\angle A = \angle C = \operatorname{arctg} 3 ; \angle B = \pi - 2 \operatorname{arctg} 3 ; \frac{1}{4}$.