Taiwan IMO Selection Camp, 2022

In each of the following solutions, the following fact is used.
Since $AC = BC = AD$, we have $\angle ABC = \angle BAC = \angle ACD = \angle ADC$. Since the quadrilaterals $APRQ$, $AQCD$ both have circumscribed circles, we obtain
$$\angle CRA = 180^\circ - \angle ARP = 180^\circ - \angle AQP = \angle DQA = \angle DCA = \angle CBA,$$
hence $A, B, C, R$ are four concyclic points; denote this circle by $\gamma$.

Solution 1. Let point $X$ be the intersection of $AQ$ and $CD$. The original problem is equivalent to proving that $B, R, X$ are collinear.
On circle $(APRQ)$ we have
$$\angle RQX = 180^\circ - \angle AQR = \angle RPA = \angle RCX$$
(where the last equality comes from $AB \parallel CD$), so $C, Q, R, X$ are four concyclic points, denoted circle $\delta$.
Using circles $\gamma$, $\delta$ we know that
$$\angle XRC = \angle XQC = 180^\circ - \angle CQA = \angle ADC = \angle BAC = 180^\circ CRB,$$

Solution 2. Denote circle $(APRQ)$ by $\alpha$. Since
$$\angle CAP = \angle ACD = \angle AQD = 180^\circ - \angle AQP,$$
we know that line $AC$ is tangent to circle $\alpha$.

Let line $AD$ meet $\alpha$ again at point $Y$; this point must lie on ray $DA$ behind $A$. Using circle $\gamma$ and the property that $AC$ is tangent to $\alpha$, we obtain
$$\angle ARY = \angle CAD = \angle ACB = \angle ARB,$$
so $Y, B, R$ are three collinear points.
Applying Pascal's theorem on the hexagon $AAYRPQ$ (here $AA$ refers to the tangent line to circle $\alpha$ passing through $A$), we obtain the intersection points of three pairs of lines
$$AA \cap RP = C, \quad AY \cap PQ = D, \quad YR \cap QA$$

Solution 3. As in Solution 1, let $X = AQ \cap CD$; we prove below that $B, R, X$ are collinear. As in Solution 2, let $\alpha = (APRQ)$, but here define point $Y$ as the second intersection point of line $BR$ with $\alpha$.
Using circle $\alpha$, and noting that line $CD$ is tangent to $\gamma$, we obtain
$$\angle RY A = \angle RPA = \angle RCX = \angle RBC. \tag{1}$$
Hence $AY \parallel BC$, and therefore $Y$ lies on line $DA$.
From (1) we also obtain $\angle RYD = \angle RCX$, so $C,D,Y,R$ are four concyclic points on a circle $\beta$. Therefore the lines $CD, AQ, YBR$ are the pairwise radical axes of the three circles $(AQCD)$, $\alpha, \beta$, so they must be concurrent.