Circumcentral midpoints Let $ABC$ be a triangle with $AB = AC$. Let $M$ be the midpoint of $BC$. Let the circles with diameters $AC$ and $BM$ intersect at points $M$ and $P$. Let $MP$ intersect $AB$ at $Q$. Let $R$ be a point on $AP$ such that $QR \parallel BP$. Prove that $CP$ bisects $\angle RCB$.
Attempt: Let $AP \cap BC=E$ and Let $D$ be orthocenter WRT $\Delta AEC$ $\implies$ $PEMD$ is cyclic. Also, $\angle EPM$ $=$ $\angle ACB$ $=$ $\angle QBE$ $\implies$ $QBEP$ & $AQPD$ are cyclic. 
 A: 
As shown above, let $E$ be the intersection of $BC$ and the extension of $AP$; join $PE$, $QE$, $GP$, and $AM$. The big idea is to show $\angle ABP=\angle BCP$ and $\angle ABP=\angle RCP$, respectively. 
We finish the proof in two steps.
Step $\mathbf 1$. Show that $\angle ABP=\angle BCP$:
Since $A,P,M,C$ are cyclic, 
$$\tag{1}\angle EPM=\angle ACB;$$
Since $AB=AC$,
$$\tag{2} \angle ACB=\angle ABC.$$
From $(1)$ and $(2)$, we conclude that $\angle EPM=\angle ABC$, and so $Q,P,E,B$ are cyclic, which then yields that 
$$\tag{3} \angle BQP=\angle AEC.$$
Since $BM$ and $AC$ are diameters, we have $BP\perp QM$ and $AP\perp PC$, and so
$$\tag{4}\angle BQP+\angle ABP=90^\circ=\angle AEC+\angle BCP.$$
$(3)$ and $(4)$ jointly imply that
$$\tag{5} \angle ABP=\angle BCP.$$
Step $\mathbf 2$: Show that $\angle ABP=\angle RCP$:
Notice that $\angle BPM=\angle APC=90^\circ$ while $\angle PMB=\angle PAC$ (since $A,C,M,P$ are cyclic), and so $\triangle PMB\sim\triangle PAC$, from which we conclude that 
$$\tag{6} \frac{PM}{AP}=\frac{BP}{PC}.$$
Next, due to the fact that $Q,P,E,B$ are cyclic (as was shown in step 1) and $\angle QPB=90^\circ$ (since $BP\perp QM$), we deduce that 
$$\tag{7}QE\perp BC.$$
Also, since $AB=AC$ while $M$ is the midpoint of $BC$, it is easy to see that 
$$\tag{8} AM\perp BC. $$
Putting $(7)$ and $(8)$ together, we get $QE\parallel AM$, and so 
$$\tag{9}\frac{QP}{PM}=\frac{QE}{AM}=\frac{BQ}{AB}.$$
Using the condition that $QR\parallel BP$, we obtain
$$\tag{10}\frac{BQ}{AB}=\frac{PR}{AP}.$$
$(9)$ and $(10)$ jointly imply that 
$$\frac{QP}{PM}=\frac{RP}{AP},$$
or equivalently, 
$$\tag{11}\frac{QP}{RP}=\frac{PM}{AP}.$$
Inspecting $(6)$ and $(11)$, one sees that 
$$\frac{QP}{RP}=\frac{BP}{PC},$$
which, together with the fact that $\angle RPC=\angle QPB=90^\circ$, yields that $\triangle PRC\sim\triangle PQB$, from which we conclude that 
$$\tag{12}\angle ABP=\angle RCP.$$
Finally, the desired result follows from $(5)$ and $(12)$. $\mathbf{Q.E.D}$
A: Here is a late alternative proof going on the path following the OP, i started typing, but the job did not let me finish it. Since the post was almost finished, here is a statement giving "all coincidences extracted from the situation". I hope this reveals "more structure" and can be used to obtain more solutions for problem.
So we claim the same, and slightly extend the results:


Let $ABC$ be a triangle with $AB = AC$.
Let $M$ be the midpoint of $BC$.
Let the circles with diameters $AC$ and $BM$ intersect at points $M$ and $P$. The circle $(BMP)$ intersects for the second time $AB$ in a point $S$. The circle $PEC$ intersects for the second time $AC$ in a point $L$.
Let $MP$ intersect $AB$ at $Q$. Let $R$ be a point on $AP$ such that $QR \parallel BP$.
Let $E$ be the intersection $AP\cap BC$.
The orthocenter $D$ of $\Delta AEC$ is in notation the intersection $AM\cap EL\cap CP$.

Prove the following statements:

*

*$QE\| AM$ and $RL\| PC$ .


*$CP$ bisects $\angle RCB$.


Proofs:
Some general facts:
We use the letter $y$ for the (measure of the) angle $\widehat {PAM}$, and try to compute all needed angles from th epicture using it together with (the measure of) $\widehat{MAC}$, which will be denoted by $\frac A2$ in the sequel. Then we have the situation from the picutre. Some small comments for the marked equal angles.

*

*The angles marked by $\displaystyle \frac A2$ are equal, the two "inside $APMC$" in $A,P$ because $APMC$ is cyclic, the two "inside $\Delta PBC$" because adding a right angle we obtain $\widehat {BPC}$, and the two in $P,R$ by correspondence w.r.t. the parallels $PQ\| PB$.

*The angles marked with $y$ in $A,C$ are equal because $APMC$ is cyclic, because of the right angles in $M,P$. The two $y$-angles in $A,B$ are equal, because adding $\widehat {BAP}$ we obtain the same $\frac 12 A$ angle (interior in $\widehat {BAM}$, respectively exterior $\widehat{BPE}$ w.r.t. the triangle $\Delta ABP$).
The two angles marked $y$ in $B,Q$ are equal because of the correspondence w.r.t. $PQ\|BP$.

*The blue angles are equal (in measure), because (in measure)
$$
\begin{aligned}
\widehat{AEC}
&=
180^\circ
-
\widehat{EAC}
-
\widehat{ACE}
=
180^\circ
-
(y+\frac {\hat A}2)-\hat C
\\
&=
180^\circ-90^\circ
- y
=
\\
&=
180^\circ-(90^\circ + y)
=
180^\circ-\widehat{PQA}
\\
&=
\widehat{BQM}
\ .
\end{aligned}
$$

*So the quadrilateral $BEPQ$ is cyclic. So we can finally also identify the angle in $E$ in this quadrilateral as being $\widehat {PEQ}=\widehat {PBQ}=y$. This gives
$$QE\|AM $$
alfter comparing the green interior angles built with the secant line $AE$.

*In particulat $QE$ is a height in $\Delta QBM$, and the heights $QE$, $BP$, $MS$ are concurrent.

*$CP$, $AM$, $EL$ are the heights in $\Delta AEC$. Let $D$ be their intersection, the orthocenter of this triangle. We show that $A,Q,P,D,L$ are on a circle. First, $ALDP$ cyclic because of the right angles in $L,P$. Then the blue angle in $Q$ is equal to the blue angle in $L$ (because of the equality of each one with the blue angle in $E$), so $AQPL$ cyclic. In particular $\widehat{PLD}=\widehat{PAD}=y$.
(Or we can extract this from $(PECL)$.)

*The power of $A$ computed w.r.t. two circles gives the equality:
$$
AL\cdot AC= AP\cdot AE = AQ\cdot AB\ .
$$
But $AB=AC$, so $AQ=AL$ and thus $QL\|BC$.

*By the homothety / similitude transformation centered in $A$ which maps $\Delta ABC$ in $\Delta AQL$ the point $P\in AE$ corresponds to the point $R\in AE$ because $BP\|QR$. This gives then $CP\| LR$, and $LR\perp AE$, and $\widehat {LRQ}=\widehat {PCB}=y$.


First proof: The idea is to use metric relations. (The following trigonometrical relations can be rephrased using similar triangles. I am not doing this, since the second proof will be synthetical.)

*

*All the above steps are dancing around the problem, but at some point we need to introduce seriously $R$ and either the connection via $C$ using angles, or using metric relations. I decided to do the last.
We will need the angle
$\widehat{EQP} =
\widehat{EQM} =
\widehat{QMA} =
90^\circ-\widehat{PMB} =
\widehat{PBM} =
\hat B-y
=90^\circ-\frac A2-y
=90^\circ-\left(\frac A2+y\right)$. The sine theorem gives using e.g. $QP$ as an intermediate
$$
\begin{aligned}
\frac{PR}{PE}
&=
\frac{PR}{PQ}
\cdot
\frac{PQ}{PE}
\cdot 
\frac{AC^2}{AC^2}
=
\frac{1}{\sin\widehat{QRP}}
\cdot
\frac{\sin\widehat{PEQ}}{\sin\widehat{EQP}}
\cdot 
\frac{AC^2}{AC^2}
\\
&=
\frac{1}{\sin(A/2)}
\cdot
\frac{\sin y}{\cos\left(\frac A2+y\right)}
\cdot 
\frac{AC^2}{AC^2}
\\
&=
\frac{AC^2\sin y}{MC\cdot AP}
=
\frac{AC^2}{MC}
\cdot
\frac{\sin \widehat {BPA}}{AB}
=
\frac{AC\sin(A/2)}{MC}
\\
&=1\ .
\end{aligned}
$$

*So $PR=PE$, and the height $CP$ in the triangle $\Delta CRE$ is also the median, so this triangle is isosceles, so we can finally also draw a green angle of measure $y$ in $\widehat {RCP}$.

$\square$

Second proof: The idea is that the whole figure has $AM$ as a symmetry, so there correspond $A\leftrightarrow A$, $B\leftrightarrow C$, $Q\leftrightarrow L$. And the question is, what is corresponding to $D$?
Well, $D$ is on the symmetry axis, so $D\leftrightarrow D$, so $D$ is the mid point of $EL$. (Draw the rectangle $EQL?$ if this is not a one second blitz.) So $D$ is the center of the circle $(EQRL)$, in particular, from $\color{blue}{DE}=DQ=\color{blue}{DR}=DL$ we need only the equality of the lengths of the) blue segments to get $\Delta DRE$ isosceles. So $PR=PE$, and then also $\Delta CRE$ isosceles.
$\square$
