Tricky circle geometry question. I've spent a while working through this question
**Let $ABC$ be a triangle. A line parallel to $BC$ meets the side $AB$ at $P$ and the side $AC$ at $Q$. The line through $C$ that is parallel to AB meets the line $PQ$ at $R$. Let $D$ be the reflection of $C$ in the line $BR$.
Prove that $D$ lies on the circumcircle of triangle APQ if and only if AB=BC.**
I can prove that if $AB=BC$ then $APQD$ are concyclic easily enough, the hard part is trying to prove the converse case to meet the if and only if condition.
I begin by assuming that $APQD$ are concyclic and try to work backwards to prove that $AB=BC$ but I keep running into roadblocks or going around in circles.
Can anyone see how to solve this?
 A: 
As shown above, join $AD$, $PD$, $BD$, $DQ$, $DF$, and $DR$. Moreover, we let $F=BR\cap AC$. We argue in several steps.
Step $\bf 1$. $\triangle BPR\cong\triangle RDB$

Since $D$ is the reflection of $C$ in $BR$, we must have $DR=RC$ and
  $BD=BC$. Since $PR\parallel BC$ and $CR\parallel AB$, the
  quadrilateral $BPRC$ is a parallelogram, and so $RC=BP$ and $PR=BC$.
  Together, $PR=BD$, and $BP=DR$, and so $\triangle BPR\cong\triangle RDB$.

Step $\bf 2$. $\triangle PBD\cong\triangle DRP$

Since $\triangle BPR\cong\triangle RDB$ (step 1), we must have
  $BD=RP$, which, together with the fact that $DR=PB$, yields the
  desired result.

Step $\bf 3$. $PD\parallel  BR$

Since $\triangle BPR\cong\triangle RDB$ (by step 1), $S_{\triangle
BPR}=S_{\triangle RDB}$. Since the two triangles share the base, they
  must have the same height, and so the distance of $P$ to $BR$ must be
  the same as that of $D$ to $BR$, which implies that $PD\parallel BR$.

Step $\bf 4$. $D,R,F,Q$ are concyclic

We argue this by showing $\angle AQD=\angle DRF$. To this end, note
  that $\angle AQD=\angle APD$ since $A,P,Q,D$ are concyclic; $\angle
APD=\angle ABR$ since $PD\parallel BR$ (by step 3); and $\angle
ABR=\angle DRF$ since $\triangle BPR\cong\triangle RDB$ (by step 1).
  Done.

Step $\bf 5$. $\triangle ABD\sim\triangle QRD$

Since $\triangle BPD\cong\triangle RDP$ (step 2), $\angle DRQ=\angle ABD$;
  since $A,D,Q,P$ are concyclic, $\angle DQR=\angle BAD$. Together, the
  desired result follows.

Step $\bf 6$. $DR=QR$

We show this by proving $\angle QDR=\angle DQR$. First note that
  $\angle DQR=\angle PDQ+\angle DPQ$ and $\angle QDR=\angle QDF+\angle
FDR$. Second, since $D,R,F,Q$ are concyclic (step 4), $\angle
QDF=\angle PRF$; since $PD\parallel PR$ (step 3), $\angle PRF=\angle
DPQ$; So we are left with showing that $\angle FDR=\angle PDQ$. To
  this end, notice the following:
$A,D,Q,P$ are concyclic $\Rightarrow$ $\angle PDQ=\angle PAQ$; 
$AB\parallel CR$ $\Rightarrow$ $\angle PAQ=\angle RCF$; 
$BR$ is the perpendicular bisector of $CD$ $\Rightarrow$ $\triangle RDF\cong\triangle RCF$ $\Rightarrow$ $\angle RCF=\angle FDR$.

Step $\bf 7$. $AB=BC$

Since $DR=QR$ (step 6) and $\triangle ABD\sim\triangle QRD$ (step 5),
  $AB=BD$. Since $BR$ is the perpendicular bisector of $CD$, $BD=BC$.
  Together, we are done.

