74th 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 $TQ=QP=2EP=EB=CP$.