Question 96
Let $P$ be a point in the interior of a triangle $ABC$ that is not on the median belonging to $A$ and satisfying $\angle CAP = \angle BCP$. Let $BP \cap CA = \{B'\}$ and $CP \cap AB = \{C'\}$. Let $Q$ be the second point of intersection of the line $AP$ and the circumcircle of $ABC$, $R$ be the point of intersection of the lines $B'Q$ and $CC'$, and $S$ be the point of intersection of the line $B'Q$ and the line passing through $P$ parallel to $AC$. Assume that the lines $B'C'$ and $QB$ intersect at a point $T$ that lies on the opposite side of the line $AB$ from $C$. Show that $\angle BAT = \angle BB'Q$ if and only if $SQ = RB'$.