Problem 1097

Solution. We will show that $M O P D$ is a parallelogram. From this it follows that $M, N$, $P$ are collinear.

Since $\angle B A D=\angle C A O=90^{\circ}-\angle A B C, D$ is the foot of the perpendicular from $A$ to side $B C$. Since $M$ is the midpoint of the line segment $B E$, we have $B M=M E=M D$ and hence $\angle M D E=\angle M E D=\angle A C B$.

Let the line $M D$ intersect the line $A C$ at $D_{1}$. Since $\angle A D D_{1}=\angle M D E=\angle A C D, M D$ is perpendicular to $A C$. On the other hand, since $O$ is the center of the circumcircle of triangle $A B C$ and $P$ is the midpoint of the side $A C, O P$ is perpendicular to $A C$. Therefore $M D$ and $O P$ are parallel.

Similarly, since $P$ is the midpoint of the side $A C$, we have $A P=P C=D P$ and hence $\angle P D C=\angle A C B$. Let the line $P D$ intersect the line $B E$ at $D_{2}$. Since $\angle B D D_{2}=\angle P D C=$ $\angle A C B=\angle B E D$, we conclude that $P D$ is perpendicular to $B E$. Since $M$ is the midpoint of the line segment $B E, O M$ is perpendicular to $B E$ and hence $O M$ and $P D$ are parallel.

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