8,9

In triangle $A B C$, a perpendicular line passing through the midpoint of side $A B$ intersects line $A C$ at point $M$, and a perpendicular line passing through the midpoint of side $A C$ intersects line $A B$ at point $N$. It is known that $M N=B C$ and line $M N$ is perpendicular to line $B C$. Find the angles of triangle $A B C$.
Let $BC = a; D$ and $E$ be the midpoints of $AB$ and $AC$. Points $E, M, D$, and $N$ lie on a circle with diameter $MN$ (since $\angle MEN = \angle MDN = 90^\circ$), $MN = a$, and $ED = \frac{\frac{a}{2}}{2}$ (the midline of triangle $ABC$). Then $DE = MN \sin \angle DME$. Therefore,

$$
\sin \angle DME = \frac{DE}{MN} = \frac{\frac{a}{2}}{a} = \frac{1}{2}
$$

Thus, either $\angle DME = 30^\circ$ or $\angle DME = 150^\circ$. Let $\angle DME = 30^\circ$ (Fig. 1). Then

$$
\angle BAC = \angle MAD = 90^\circ - \angle EMD = 60^\circ.
$$

The inscribed angles $EDN$ and $EMN$ subtend the same arc, $DE \parallel BC$ (as the midline), and $\angle ABC = \angle DMN$ (as angles with correspondingly perpendicular sides). Therefore,

$$
\angle EMN = \angle EDN = \angle EDA = \angle ABC = \angle DMN
$$

Thus,

$$
\angle ABC = \frac{1}{2} \angle EMD = 15^\circ.
$$

Then

$$
\angle ACB = 180^\circ - 60^\circ - 15^\circ = 105^\circ
$$

If $\angle DME = 150^\circ$ (Fig. 2), reasoning similarly, we get that

$$
\angle BAC = 60^\circ, \angle ACB = 15^\circ, \angle ABC = 105^\circ.
$$

## Answer

$\angle A = 60^\circ, \angle B = 15^\circ, \angle C = 105^\circ$ or $\angle A = 60^\circ, \angle B = 105^\circ, \angle C = 15^\circ$.