Question 53
Let $ABC$ be an acute triangle such that $AB \neq AC$, with circumscribed circle $\Gamma$ and circumcenter $O$. Let $M$ be the midpoint of $BC$ and $D$ be a point on $\Gamma$ such that $AD \perp BC$. Let $T$ be a point such that $BDCT$ is a parallelogram and $Q$ a point on the same side of $BC$ as $A$, such that
$$
\angle BQM = \angle BCA \text{ and } \angle CQM = \angle CBA.
$$
Let the line $AO$ intersect $\Gamma$ at $E$, ($E \neq A$) and let the circumscribed circle of $\triangle ETQ$ intersect $\Gamma$ at point $X \neq E$. Prove that the points $A, M$ and $X$ are collinear.
$$
\angle BQM = \angle BCA \text{ and } \angle CQM = \angle CBA.
$$
Let the line $AO$ intersect $\Gamma$ at $E$, ($E \neq A$) and let the circumscribed circle of $\triangle ETQ$ intersect $\Gamma$ at point $X \neq E$. Prove that the points $A, M$ and $X$ are collinear.