TITLE: Find the angle or prove it is constant
QUESTION [7 upvotes]: 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?

REPLY [2 votes]: 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 $\angle AQR = \angle ABR = 36^0$.