As $M$ is the midpoint of $AP$ and $R$ is the midpoint of $BC$, we have the vector equation $\vec{RM} = \frac{\vec{CA} + \vec{BP}}{2}$. Now look at triangle $BEC$. Since vectors $\vec{CA}$ and $\vec{BP}$ have the same magnitude, and their directions are along the sides of the triangle, their vector sum must be parallel to the internal angle bisector of $\angle BEC$. This means $\vec{RM}$ is perpendicular to the external angle bisector of $\angle BEC$, which is $\angle BEA$, as desired.

@ above - that's clever! My way: - Let $RM$ intersect $BE$ at $Y$ and $CA$ at $Z$. it is enough to prove that $EY=EZ$. To do so is just two applications of menelaeus, taking $YM$ as the traversal of triangle $APE$ and then $RY$ as the traversal of triangle $BEC$. It made me laugh how it all suddenly drops out.

I don’t see how you can conclude with that so directly. Could you explain more?

Let $Q$ be the reflection of $P$ with respect to $R$. Then $BPCQ$ is a parallelogram, so $\angle BEA = \angle ACQ$, and $MR \parallel AQ$ is a midsegment in triangle $APQ$, so it remains to show that the bisector of $\angle ACQ$ is perpendicular to $AQ$. But this holds since $AC = BP = CQ$, done.

Reflect $P$ over $R$ to get $P'$. First of all by homothety $MR\parallel AP'$. Further $BP\parallel CP'$ as $BPCP'$ is a parallelogram. Hence the bisector of $\angle AEB$ is parallel to the bisector of $\angle ACP'$. Yet $AC=BP=CP'$ hence $\triangle ACP'$ is isoscles, and it is well known the angle bisector is perpendicular to the opposite side.

Actually, this is isomorphic to the following well known result (just take $P$ to be outside of $ABC$ to see is properly): if quadrilateral $ABCD$ is such that $AB = CD$, then with $E = AB \cap CD$ the line through the midpoints of $AD$ and $BC$ is parallel to the angle bisector of $\angle BEC$.