1) Triangles $PMD$ and $PDN$ are similar (indeed, $\widehat{PDM}$ is half inscribed, $\widehat{DNP}$ is inscribed and both subtend the same arch in the Euler circle; moreover $\hat{P}$ is the same in both triangles). Therefore
$$\frac{PM}{PD} = \frac{DM}{DN} = \frac{PD}{PN} \implies \frac{PM}{PN} = \left(\frac{DM}{DN}\right)^2. \quad (*)$$

Points $M$, $N$ and $B$ belong to a circle of center $A'$ and radius $a/2$. (Indeed, in the triangle $A'BN$ we have $A'B = a/2$; $BN = a \cos B$; and by the cosine law
$$A'N^2 = A'B^2 + BN^2 - 2 A'B \cdot BN \cdot \cos B = \frac{a^2}{4}.$$
In the triangle $A'MC$ we repeat the argument and we have $A'M = a/2$).

Then, being $MA' = NA' = a/2$, the angle $\angle MDN$ has an angle bisector $DA'$, and so, if we call $T = MN \cap AA'$, results
$$\frac{DM}{DN} = \frac{TM}{TN}.$$

But $MN$ is antiparallel of $BC$ with respect to $AB$ and $AC$, and so
$$\frac{TM}{TN} = \frac{AM^2}{AN^2} = \frac{c^2}{b^2}. \qquad (**)$$

Combining these results, we have $\frac{PM}{PN} = \frac{c^4}{b^4}$; in a similar way we get
$$\frac{QN}{QL} = \frac{a^4}{c^4} \quad \text{and} \quad \frac{RL}{RM} = \frac{b^4}{a^4},$$
therefore
$$\frac{PM \cdot QN \cdot RL}{PN \cdot QL \cdot RM} = 1.$$
By the converse of the Menelaus theorem, the points $P$, $Q$ and $R$ are collinear.