Turkish Mathematical Olympiad

Since $PS \parallel AC$ and $\angle CAP = \angle BCP$ we have $\angle QPC = \angle ACB = \angle AQB = \angle PQB$. Therefore $BQ \parallel PC$.

We will first show that $SQ = RB'$ if and only if $AB \parallel B'Q$. Since $PS$ and $AB'$ are parallel, we have $\frac{PQ}{PA} = \frac{SQ}{SB'}$. If $SQ = RB'$, then $\frac{PQ}{PA} = \frac{RB'}{RQ} = \frac{PB'}{PB}$ as $BQ \parallel PC$, and hence $AB \parallel B'Q$. On the other hand, if $AB \parallel B'Q$, then using $BQ \parallel PC$ we obtain $\frac{RB'}{RQ} = \frac{PB'}{PB} = \frac{PQ}{PA} = \frac{SQ}{SB'}$, and hence $SQ = RB'$.

Now we will show that $AB \parallel B'Q$ if and only if $\angle BAT = \angle BB'Q$. Let $AP \cap B'C' = \{U\}$ and let $D$ be the second point of intersection of the line $AP$ and the circumcircle of the triangle $C'PB'$. Using $PC' \parallel QT$ we obtain $\frac{UD}{UB'} = \frac{UC'}{UP} = \frac{UT}{UQ}$

and conclude that the triangles $TDC'$ and $QB'P$ are similar. In particular, $\angle PB'Q = \angle C'DT$ and $\angle DTC' = \angle B'QP$.

We will show that $D$ cannot coincide with $A$. Suppose it does. Then $A, C', P, B'$ are concyclic and hence $C'B'$ and $BC$ are parallel. Let $A'$ be the point of intersection of $AP$ and $BC$. It follows that the triangles $PBA'$ and $BAA'$ are similar, and therefore $A'B^2 = A'P \cdot A'A$. Similarly we obtain $A'C^2 = A'P \cdot A'A$ and conclude that $A'B = A'C$, which is a contradiction as $P$ is not on the median belonging to $A$.

Let us assume that $D$ is between $A$ and $U$. A similar argument works if $A$ is between $D$ and $U$. If $\angle BAT = \angle BB'Q$, then $\angle C'AT = \angle BAT = \angle BB'Q = \angle PB'Q = \angle C'DT$, and $T, A, D, C'$ are concyclic. Therefore $\angle PAB = \angle DAC' = \angle DTC' = \angle B'QP$ and hence $AB \parallel B'Q$. On the other hand if $AB \parallel B'Q$, then $\angle DTC' = \angle B'QP = \angle DAC'$, and $T, A, D, C'$ are concyclic. Therefore $\angle BAT = \angle C'AT = \angle C'DT = \angle PB'Q = \angle BB'Q$.