Question 43
In the circumcircle of triangle $ABC$, let $A_1$ be the point diametrically opposite to the vertex $A$. Let $A'$ be the point where the line $AA_1$ meets the side $BC$. The perpendicular to the line $AA'$ drawn through $A'$ meets the sides $AB$ and $AC$ (or their extensions) at $M$ and $N$, respectively. Prove that the points $A$, $M$, $A_1$ and $N$ lie on a circle whose center lies on the altitude from $A$ of triangle $ABC$.