APMO

To show $B$, $E$, $R$ are collinear, it is equivalent to show the lines $AD$, $BE$, $CQ$ are concurrent. Let $CQ$ intersect $AD$ at $R$ and $BE$ intersect $AD$ at $R^{\prime }$. We shall show $RD/RA=R^{\prime }D/R^{\prime }A$ so that $R=R^{\prime }$.

Since $△PAD$ is similar to $△PDC$ and $△PAB$ is similar to $△PBC$, we have $AD/DC=PA/PD=PA/PB=AB/BC$. Hence, $AB\cdot DC=BC\cdot AD$. By Ptolemy's theorem, $AB\cdot DC=BC\cdot AD=\frac{1}{2}CA\cdot DB$. Similarly $CA\cdot ED=CE\cdot AD=\frac{1}{2}AE\cdot DC$.

Thus $$\frac{DB}{AB}=\frac{2DC}{CA},\text{\hspace{1em}(3)}$$ and $$\frac{DC}{CA}=\frac{2ED}{AE}\text{\hspace{1em}(4)}$$ Since the triangles $RDC$ and $RCA$ are similar, we have $\frac{RD}{RC}=\frac{DC}{CA}=\frac{RC}{RA}$. Thus using (4) $$\frac{RD}{RA}=\frac{RD\cdot RA}{R{A}^2}={\left(\frac{RC}{RA}\right)}^2={\left(\frac{DC}{CA}\right)}^2={\left(\frac{2ED}{AE}\right)}^2\text{\hspace{1em}(5)}$$ Using the similar triangles $AB{R}^{\prime }$ and $ED{R}^{\prime }$, we have $R^{\prime }D/R^{\prime }B=ED/AB$. Using the similar triangles $DB{R}^{\prime }$ and $EA{R}^{\prime }$ we have $R^{\prime }A/R^{\prime }B=EA/DB$. Thus using (3) and (4), $$\frac{R^{\prime }D}{R^{\prime }A}=\frac{ED\cdot DB}{EA\cdot AB}={\left(\frac{2ED}{AE}\right)}^2\text{\hspace{1em}(6)}$$ It follows from (5) and (6) that $R=R^{\prime }$.