Romanian Mathematical Olympiad

a) Let $F$ be the intersection of the lines $CP$ and $AB$ and $E$ be the foot of the perpendicular from $A$ to $BM$.

Since $\angle FCB = \angle MBA = 90^\circ - \angle CBM$ and $CB = BA$, the right triangles $CBF$ and $BAM$ are congruent, whence $FB = MA = \frac{AB}{2}$, so $F$ is the midpoint of side $AB$.
We deduce that $FP$ is a midsegment in the triangle $AEB$, so $BP = EP$ and the congruence of the triangles $CPB$ and $BEA$ yields $PB = AE$ and $CP = BE$.
The triangle $EAP$ is right and isosceles, so $\angle APE = 45^\circ$, thus the triangle $APQ$ is also an isosceles right triangle. Consequently, the altitude $AE$ is also a median.
We obtain $QP = 2EP = EB = PC$, hence the triangle $PCQ$ is also an isosceles right triangle, from which follows that $\angle PCQ = 45^\circ$ and $CQ \parallel AP$.

b) Since $CD = CB$, $\angle DCP = \angle CBE$ and $PC = EB$, the triangles $DPC$ and $CEB$ are congruent, whence $DP = CE = CB = DA$.
From the congruence of the triangles $DEA$ and $DEP$ (SSS), it follows that $\angle ADE = \angle PDE$, so $DE$ is the supporting line of the bisector and the altitude in the isosceles triangle $DAP$.
Since $DE \perp QC$, it follows that $CQ$ is the perpendicular bisector of the side $DE$ in the isosceles triangle $CDE$, thus $QD = QE = \frac{QP}{2} = \frac{PC}{2}$.
Therefore $DQ$ is a midsegment in the triangle $TPC$, whence $TQ = QP = 2EP = EB = CP$.