Find the angle or prove it is constant We are given a regular pentagon ABCDE (with the letters in clockwise sequence).
A point Q moves on side BC and we draw a circle with center Q which passes through the vertex A and intersects the diagonal BE in point R. Prove that angle AQR is constant.

If $s$ is the side of the regular pentagon, then the length of its diagonal is $d = s*\frac {\sqrt5+1}{2}$.
Then, length of segment RE is $d - s = s*\frac {\sqrt5-1}{2}$.
Also if Q coincides with B, then the triangle ABR is a "golden" triangle and its base side $t$ is $\frac {s}{φ}$ where $φ$ is the golden ratio. So $t = RE$ and the angle $ARE = 180^\circ-72^\circ = 108^\circ$. So AR and RE are 2 sides of a regular pentagon. So, angle $AQR = 36^\circ$.
Edit: I added the diagram, as you asked.
The pink polygon is not part of the problem - I just sketched it in my attempt to prove the question.
I also tried to prove that quadrilateral $ABQR$ is cyclic, in which case, angle $α$ would be equal to $ABE$, which we know is $36^\circ$.
Is this a sufficient proof?
 A: 
Let $|AB|=1$, $|BQ|=t\in[0,1]$.
Then $|AC|=|BE|=\tfrac{1+\sqrt5}2$,
$|CQ|=1-t$,
\begin{align} 
\triangle AQB:\quad
|AQ|^2=|QP|^2
&=t^2+\tfrac12(\sqrt5-1)\,t+1
\tag{1}\label{1}
,\\
\triangle AEP:\quad
|AP|^2&=1+|EP|^2-\tfrac12|EP|(1+\sqrt5)
\tag{2}\label{2}
,\\
\triangle APQ:\quad
|AP|^2&=
(2\,|AQ|\sin\tfrac\phi2)^2
\tag{3}\label{3}
,\\
\triangle EQB:\quad
|EQ|^2&=
t^2-t+\tfrac32+\tfrac12\sqrt5
\tag{4}\label{4}
.
\end{align}
Using  Stewart’s Theorem
wrt $\triangle EQB$,
we get
\begin{align} 
|EP|&=
\tfrac12(1-\sqrt5)\,t-\tfrac12+\tfrac12\sqrt5
\tag{6}\label{6}
.
\end{align}
Then it follows from \eqref{2} and \eqref{3}
that
\begin{align}
\cos\phi&=\tfrac14+\tfrac14\sqrt5
,\\
\phi&=36^\circ
.
\end{align}
A: Center of the circle $Q$ moves on $BC$. So when the center is at $B$, we draw a circle passing through $A$, in other words, the radius of the circle is $AB$. We now extend $EB$ and say it meets the circle at $H$.

Now the first thing to note is that $\angle CBH = \angle ABC = 108^0 \ $
Also, $BH = AB$ as $B$ is the center of the circle.
As $BC$ is the angle bisector of $\angle ABH$ and $AB = BH$, for any point $Q$ on $BC$, we have $QH = QA = QR$.
So it follows that $\angle BAQ = \angle BHQ = \angle BRQ$.
That shows quadrilateral $BARQ$ is cyclic.
Hence $\angle AQR = \angle ABR = 36^0$.
A: Choose point $S$ on $FE$ (see figure below) such that $\angle DAS=\angle BAQ$ (this is possible for any position of $Q$ on $BC$). I'll show later that $QS=QA$, so that $S$ is the same as point $R$ defined in the question. But this alternate definition makes the proof much easier.
In fact, note that triangles $QBA$ and $SFA$ are similar by construction, because $\angle ABQ=\angle AFS=108°$ (AA criterion of similarity), hence $AS:AQ=AF:AB$. From this it follows that triangles $ASQ$ and $AFB$ are similar too, because
$\angle SAQ=\angle FAB$ (SAS criterion of similarity).
It follows that triangle $ASQ$ has the same angles as $AFB$ and is in particular isosceles. Thus we get $QS=QA$, that is $S=R$, and $\angle AQR=\angle AQS=\angle ABF=36°$.

