Irish Mathematical Olympiad

Step I: Reduce to the case when $M$ coincides with $B$. The following provides two versions of this step:

Version 1: The line through $B$ parallel to $MN$ intersects $AP$ at a point $E$ and $AC$ at a point $Q$. From the similar triangles $AMD$ and $ABE$ it follows that
$$\frac{|MD|}{|BE|} = \frac{|AD|}{|AE|}$$
and from the similar triangles $ADN$ and $AEQ$ it follows that
$$\frac{|ND|}{|QE|} = \frac{|AD|}{|AE|}$$
Thus $|AD| = |DM| = 2|DN|$ is equivalent to $|AE| = |EB| = 2|EQ|$.

Version 2: Let $B'C'$ be a line parallel to $BC$, with $B' = M$ and $C'$ on $AC$. Let $P'$ be the intersection point of $B'C'$ with $AP$. From the similar triangles $AB'P'$ and $ABP$ it follows that
$$\frac{|B'P'|}{|BP|} = \frac{|AP'|}{|AP|}$$
and from the similar triangles $AC'P'$ and $ACP$ it follows that
$$\frac{|C'P'|}{|CP|} = \frac{|AP'|}{|AP|}$$
Thus $P$ is the midpoint of $BC$ if and only if $P'$ is the midpoint of $B'C'$. Also, since the triangles $AB'C'$ and $ABC$ are similar, then $|AC| = |BC|$ is equivalent to $|A'C'| = |B'C'|$.

Step II: With the notations from Step I, Version 1, prove that $Q$ is the midpoint of $AC$ and $E$ is the centroid of the triangle $ABC$. The following provides two versions of this step:

Version 1: Menelaus' theorem for $\triangle BQC$ with secant line $AP$ gives
$$\frac{|QE|}{|BE|} \cdot \frac{|BP|}{|CP|} \cdot \frac{|CA|}{|QA|} = 1 .$$
Because $|BE| = 2|QE|$ and $|BP| = |CP|$, it follows that $|CA| = 2|QA|$ and so $Q$ is the midpoint of $AC$. Hence, $E$ is the centroid of $\triangle ABC$.

Version 2: Let $Q'$ denote the midpoint of $AC$, and let $E'$ be the intersection of $BQ'$ with $AP$. We will prove $Q = Q'$ using proof by contradiction. Assume they are different. The triangles $Q'PC$ and $ABC$ are similar since
$$\frac{|CQ'|}{|AC|} = \frac{|CP|}{|BC|} = \frac{1}{2}$$
and so $|AB| = 2|PQ'|$ and also $PQ'$ is parallel to $AB$. Thus the triangles $AE'B$ and $PE'Q'$ are similar which implies
$$\frac{|AB|}{|PQ'|} = 2 = \frac{|BE'|}{|E'Q'|} = \frac{|BE|}{|EQ|}$$
But this would imply that $EE'$ is parallel to $QQ'$ which is a contradiction, since they meet at $A$.

Step III: Because $E$ is the centroid of $\triangle ABC$, we have $|AE| = 2|EP|$ and $|BE| = 2|EQ|$. Because $|AE| = 2|EQ|$, this implies $|AE| = |BE|$ and $|EQ| = |EP|$. Therefore, the triangles $AEQ$ and $BEP$ are congruent, thus
$$$$
|BC| = 2|BP| = 2|AQ| = |AC| .