I found this problem on a fb group and the author asked to use Miquel's theorem to solve it:
let $ABCD$ be a quadrilateral with $AB=AD$, and suppose $\angle BCD=\tfrac{1}{2}\bigl(180^\circ-\angle BAD\bigr)$.
Let $P$ be any point on the interior of the side $CD$. Let $Q=AP\cap BC$ and $R=DQ\cap BP$.
Prove that $\angle BRD=\angle BCD$.
I tried to use that theorem to solve the problem, but found no way. Do you know how?
The claim is equivalent to $R\in\Gamma_{BCD}$.
The circle centered at $A$ through $B$ is orthogonal to $\Gamma_{BCD}$, so $A$ is the pole of $BD$ with respect to $\Gamma_{BCD}$ and we may rephrase the problem in the following way:
Equivalent theorem. If $DBCR$ is a cyclic quadrilateral with diagonals meeting at $P$ and $Q=BC\cap RD$, then the pole of $BD$ with respect to the circumcircle lies on the $PQ$ line.
Proof. Let $S=BD\cap CR$ be the missing vertex of the complete quadrilateral. Then $SQ=\text{pol}(P)$ and by De La Hire's theorem $S=\text{pol}(QP)$ and $Q=\text{pol}(SP)$. Since $QS,PS,BD$ are concurrent, their poles are collinear. In other terms, $A\in PQ=\text{pol}(S)$ since $S\in\text{pol}(A)$.
Let $K$ be the intersection of the circle through $A,B,D$ with the line $AP$ (it is easy to see that such a point always exists).
Then the quadrilateral $ABKD$ is cyclic. Note that $\angle BDA = \angle DBA=\frac{1}{2}(180-\angle BAD)$, (since the triangle is isosceles), so $\angle BDA = \angle AKB = \angle DKA $.
Since $\angle PCB = \angle PKB$, the quadrilateral $BCKP$ is cyclic.
Also, because $\angle DCQ = \angle DKQ$, the quadrilateral $QCKD$ is cyclic.
By Miquel's theorem, there exists a unique point through which the circles $PBC$, $QDC$, and $RDP$ all intersect.
The points of intersection of $QDC$ and $PBC$ are $C$ and $K$, so one of them must also lie on $RDP$.
It cannot be $C$, because in that case we would have three collinear points on the same circle, which is impossible.
Therefore it must be $K$, and hence $RDPK$ is cyclic.
Thus $$ \angle DRP = \angle DKP = \angle DCB. $$
