Asian Pacific Mathematics Olympiad

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 $R'$. We shall show $RD / RA = R'D / R'A$ so that $R = R'$.

Since $\triangle PAD$ is similar to $\triangle PDC$ and $\triangle PAB$ is similar to $\triangle PBC$, 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
$$\frac{DB}{AB} = \frac{2DC}{CA}, \tag{3}$$
and
$$\frac{DC}{CA} = \frac{2ED}{AE} \tag{4}$$

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