Argentine National Olympiad 2015

Let $M$ be the midpoint of $BC$, and let line $DM$ intersect



line $AB$ at $Q$ (it is exterior to the segment $AB$). Then $BQM=CDM$ as $AB\parallel CD$. On the other hand $CDM=PDM$ by hypothesis (DM is the bisector of $CDP$). So $PQD=PDC$ and hence $PQ=PD$. In addition $BQ=CD=3$ because triangles $BQM$ and $CDM$ are congruent ($BM=CM$, $M\hat{B}Q=M\hat{C}D=90^{\circ }$, $BQM=CDM$).

Set $BP=x$. Then $PQ=PB+BQ=x+3$ and $PD=PQ=x+3$. Apply Pythagoras theorem to triangle $PDA$ in which $AP=3-x$, $AD=2$, $PD=x+3$. This gives $$(3-x{)}^2+2^2=(3+x{)}^2,$$ and we find $x=\frac{1}{3}$.