The quadrilaterals $\mathrm{O}_{3} M O_{2} A_{1}, \mathrm{O}_{3} M O_{1} A_{2}$ and $O_{1} M O_{2} A_{3}$ are rhombuses. Therefore, $O_{2} A_{1} \| M O_{3}$ and $M O_{3} \| O_{1} A_{2}$, which imply $O_{2} A_{1} \| O_{1} A_{2}$. Because $O_{2} A_{1}=O_{3}^{*} M=O_{1} A_{2}$ the quadrilateral $O_{2} A_{1} A_{2} O_{1}$ is a parallelogram and then $A_{1} A_{2} \| O_{1} O_{2}$ and $A_{1} A_{2}=O_{1} O_{2}$. Similarly, $A_{2} A_{3} \| O_{2} O_{3}$ and $A_{2} A_{3}=O_{2} O_{3} ; A_{3} A_{1} \| O_{3} O_{1}$ and $A_{3} A_{1}=O_{3} O_{1}$. The triangles $A_{1} A_{2} A_{3}$ and $\mathrm{O}_{1} \mathrm{O}_{2} \mathrm{O}_{3}$ are congruent.

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Since $A_{3} M \perp O_{1} O_{2}$ and $O_{1} O_{2} \| A_{1} A_{2}$ we infer $A_{3} M \perp A_{1} A_{2}$. Similarly, $A_{2} M \perp A_{1} A_{3}$ and $A_{1} M \perp A_{2} A_{3}$. Thus, $M$ is the orthocenter for the triangle $A_{1} A_{2} A_{3}$.

## GEO.6.

Consider an isosceles triangle $A B C$ with $A B=A C$. A semicircle of diameter $E F$, lying on the side $B C$, is tangent to the lines $A B$ and $A C$ at $M$ and $N$, respectively. The line $A E$ intersects again the semicircle at point $P$.

Prove that the line $PF$ passes through the midpoint of the chord $M N$.

Solution. Let $O$ be the center of the semicircle and let $R$ be the midpoint of $M N$. It is obvious that $M N$ is perpendicular to $A O$ at point $R$. Since $\angle A N O$ is right, then from the leg theorem we have $A N^{2}=A R \cdot A O$. From the power of a point theorem,

$$A P \cdot A E=A N^{2}=A M^{2}=A R \cdot A O$$

Using the same theorem we infer that points $P, R, O$ and $E$ are concyclic, hence $\angle R P E$ is right. As $\angle F P E$ is also a right angle, the conclusion follows.