IMO Shortlist, 2021

Common remarks. The introductory steps presented here are used in all solutions below.
Since $A C = B C = A D$, we have $\angle A B C = \angle B A C = \angle A C D = \angle A D C$. Since the quadrilaterals $A P R Q$ and $A Q C D$ are cyclic, we obtain
$$\angle C R A = 180^{\circ} - \angle A R P = 180^{\circ} - \angle A Q P = \angle D Q A = \angle D C A = \angle C B A,$$
so the points $A, B, C$, and $R$ lie on some circle $\gamma$.

Solution 1. Introduce the point $X = A Q \cap C D$; we need to prove that $B, R$ and $X$ are collinear.
By means of the circle $(A P R Q)$ we have
$$\angle R Q X = 180^{\circ} - \angle A Q R = \angle R P A = \angle R C X$$
(the last equality holds in view of $A B \parallel C D$), which means that the points $C, Q, R$, and $X$ also lie on some circle $\delta$.
Using the circles $\delta$ and $\gamma$ we finally obtain
$$\angle X R C = \angle X Q C = 180^{\circ} - \angle C Q A = \angle A D C = \angle B A C = 180^{\circ} - \angle C R B,$$
that proves the desired collinearity.

Solution 2. Let $\alpha$ denote the circle $(A P R Q)$. Since
$$\angle C A P = \angle A C D = \angle A Q D = 180^{\circ} - \angle A Q P,$$
the line $A C$ is tangent to $\alpha$.
Now, let $A D$ meet $\alpha$ again at a point $Y$ (which necessarily lies on the extension of $D A$ beyond $A$). Using the circle $\gamma$, along with the fact that $A C$ is tangent to $\alpha$, we have
$$\angle A R Y = \angle C A D = \angle A C B = \angle A R B,$$
so the points $Y, B$, and $R$ are collinear.
Applying Pascal's theorem to the hexagon $A A Y R P Q$ (where $A A$ is regarded as the tangent to $\alpha$ at $A$), we see that the points $A A \cap R P = C$, $A Y \cap P Q = D$, and $Y R \cap Q A$ are collinear. Hence the lines $C D$, $A Q$, and $B R$ are concurrent.

Solution 3. As in Solution 1, we introduce the point $X = A Q \cap C D$ and aim at proving that the points $B, R$, and $X$ are collinear. As in Solution 2, we denote $\alpha = (A P Q R)$; but now we define $Y$ to be the second meeting point of $R B$ with $\alpha$.
Using the circle $\alpha$ and noticing that $C D$ is tangent to $\gamma$, we obtain
$$\begin{equation*} \angle R Y A = \angle R P A = \angle R C X = \angle R B C . \tag{1} \end{equation*}$$
So $A Y \parallel B C$, and hence $Y$ lies on $D A$.
Now the chain of equalities (1) shows also that $\angle R Y D = \angle R C X$, which implies that the points $C, D, Y$, and $R$ lie on some circle $\beta$. Hence, the lines $C D$, $A Q$, and $Y B R$ are the pairwise radical axes of the circles $(A Q C D)$, $\alpha$, and $\beta$, so those lines are concurrent.