NMO Selection Tests for JBMO

Let $Q$ be the point in which the circumcircle of $BCD$ meets again the line $AB$. Then $\angle QEA = \angle QBC = \angle ECP$, hence $EQ \parallel PC$. Moreover, $\angle ECP = \angle ECD$ implies $QD$ is parallel to $EC$, hence $EQ = CD = CP$. It follows that $EQCP$ is a parallelogram, which leads to the conclusion.

Second solution. (Given in the contest.) If $\{O\} = BD \cap AC$, it is easy to prove that $BA$ and $BO$ are isogonal in the angle $\angle EBC$, therefore, in order to show that $BA$ is the median, one has to prove that $BO$ is the symmedian. This follows readily by computation, using Steiner's Theorem.
Indeed, $\angle EBA = \angle BAC - \angle BEA = 180^\circ - \angle ABC - \angle ACB - \angle BDC = 180^\circ - \angle ECD - \angle ECB - \angle BDC = \angle CBD$.
Triangles $OEB$ and $ODC$ are similar, hence $\frac{EO}{OD} = \frac{EB}{CD}$. (1)
Triangles $OCB$ and $ODE$ are also similar, therefore $\frac{OC}{OD} = \frac{BC}{ED}$. (2)
We have $\angle ADC = \angle APC = \angle PBC + \angle PCB = \angle ACB = \angle EDB$, which shows that the triangles $ACD$ and $EBD$ are similar. Hence $ED = \frac{EB \cdot AD}{AC}$ and, as $AD = \frac{CD \cdot AC}{BC}$ (from $\Delta ACB \sim \Delta APC \sim \Delta ADC$), it follows that $ED = \frac{EB \cdot CD}{BC}$.
From (1) and (2) we obtain $\frac{EO}{OC} = \frac{EB \cdot ED}{CD \cdot BC} = \left(\frac{EB}{BC}\right)^2$, which shows that $BO$ is indeed the symmedian.

Third solution. (Given in the contest by Andrei Mărginean.) Let $a = \angle EBA$, $x = \angle ABC$. We have that $\angle DCE = \angle ACP = x$ and $\angle EBC = a + x$. As $B$, $C$, $D$, $E$ are concyclic, it follows that $\angle CDE = 180^\circ - a - x$, $\angle DEC = 180^\circ - \angle EDC - \angle DCE = a = \angle EBA$. But $AE$ is the perpendicular bisector of $[PD]$, hence $\angle PEA = \angle DEC = \angle EBA$, which shows that the triangles $APE$ and $AEB$ are similar. It follows that $AE^2 = AB \cdot AP$. But triangles $ACP$ and $ABC$ are also similar, hence $AC^2 = AP \cdot AB = AE^2$, and the conclusion follows immediately.