Let $M, A', B', C'$ be midpoints of $AD, BC, CA, AB$ respectively. Then a homothety with ratio $1.5$ takes $Q\to B'$, $R\to C'$. Since $D\in\odot(A'B'C')$, this homothety takes $P$ to $A'$ or $DP : PA' = 2 : 1$. However, if we let $T = BQ\cap CR\cap MA'$, then Menelaus theorem on $\triangle MBA'$ and $\overline{CTR}$ gives $T$ is midpoint of $MA'$. Hence using the converse of Menelaus theorem on $\triangle MDA'$ and $\overline{ATP}$, we get the conclusion.

Using Cartesian coordinates, we have $R=\frac13(A+B+D)$ and $Q=\frac13(A+C+D)$. Since $R-Q=\frac13(B-C)$, we have $RQ\parallel BC$, so $DPQR$ is a cyclic trapezoid, whence the projection of the midpoint $\frac12(Q+R)=\frac16(2A+2D+B+C)$ onto $BC$ is exactly $\frac12(D+P)$. Since projection is a linear function, or by just letting $DA,DB$ be the coordinate axes, we see that this projection is just $$\frac16(2D+2D+B+C)=\frac12(D+P),\quad \implies\quad P=\frac13(B+C+D).$$Therefore, the point-centroid lines $AP,BQ,CR$ concur at the centroid $\frac14(A+B+C+D)$ of quadrilateral $ABCD$.

cheva sinusi

Actually, there is no need for any computation in this problem: Clearly $QR \parallel BC$ and a quick angle chase shows that $\angle PRQ=\angle CDQ=\gamma$ as well as $\angle RQP=\angle RPB=\beta$. Hence triangles $PQR$ and $ABC$ are similar with parallel sides, hence the desired point of intersection is just the center of the homothety sending one to the other.

Using Desargues's theorem we know that $AP$, $BQ$ and $CR$ are concurrent if and only if $QR\cap BC$ , $QP\cap AB$ and $PR\cap AC$ are collinear. By angle chasing these points are all on the infinity line so they are collinear.

Valencia_Ann wrote: Using Desargues's theorem we know that $AP$, $BQ$ and $CR$ are concurrent if and only if $QR\cap BC$ , $QP\cap AB$ and $PR\cap AC$ are collinear. By angle chasing these points are all on the infinity line so they are collinear. Cheers $;)$

In general, it is true that for a triangle $ABC$ and point $D$, if $P,Q,R$ are the centroids of triangles $BCD,CDA,ADB$ respectively, then $\triangle PQR$ is homothetic to $\triangle ABC$. This follows since $Q-R=\tfrac{A+D+C}{3}-\tfrac{A+D+B}{3}=\tfrac{C-B}{3}$, hence $\overline{QR} \parallel \overline{BC}$, and similarly the other pairs of sides are parallel. We apply this to the given problem. Set $\overline{BC}$ as the $x$-axis. Since $Q=\tfrac{A+C+D}{3}$ and $R=\tfrac{A+B+D}{3}$, $\overline{QR} \parallel \overline{BC}$, hence the center of $(PQRD)$ has $x$-coordinate equal to that of $\tfrac{A+D+\tfrac{B+C}{2}}{3}=\tfrac{2D+\tfrac{B+C}{2}}{3}$. Thus $P$ has $x$-coordinate equal to that of $\tfrac{D+B+C}{3}$, and since its $y$-coordinate is zero it equals $\tfrac{B+C+D}{3}$, so $P$ is the centroid of degenerate triangle $BCD$. $\blacksquare$