8.1. Let $M$ be a point on the arc $A B$ of the circumcircle of a tringle $A B C$. The perpendicular to the radius $O A$ drawn from $M$ intersects the sides $A B$, $A C$ at the points $K, L$ respectively ( $O$ is the center of the circumcircle). Similarly, the perpendicular to the radius $O B$ drawn from $M$ intersects the sides $A B, B C$ at the point $N, P$ respectively. If $K L=M N$, compute the angle $\angle M L P$.

(Greece)
8.1. Let us prove that $\overline{M K}=\overline{K L}=\overline{M N}=\overline{N P}$. From

$$
\begin{aligned}
& \angle O A B=\angle O B A=\frac{1}{2}(\pi-2 \angle A C B)=\frac{\pi}{2}-\angle A C B \\
& \angle A K L=\angle B N P=\angle A C B
\end{aligned}
$$

we obtain

$$
\overline{M K}=\overline{M N}=\overline{K L}
$$

The triangles $A K L$ and $P N B$ are congruent (they are both conguent to $A B C$, so that

$$
\overline{A K} \cdot \overline{B N}=\overline{K L} \cdot \overline{P N}
$$

The triangles $A M K$ and $M B N$ are congruent as well, therefore

$$
\overline{A K} \cdot \overline{B N}=\overline{M K} \cdot \overline{M N}
$$

From (1), (2) and (3) we have $\overline{M K}=\overline{M N}=\overline{K L}=\overline{N P}$. Therefore $K N \| P L$, and so $\angle M L P=\angle M K N=\angle A C B$.