Irish Mathematical Olympiad

Let $AM$ meet $PO$ at $F$ and let $PO$ extended meet the circle at $G$ and $H$ with $G$ between $P$ and $O$. Let $AB$ meet $PO$ at $Q$.

First note that $F$ is the centroid of $\triangle ABP$ and so $|FQ| = \frac{1}{3}|PQ|$. Using that $D$ is the mid-point of $PO$ and denoting the radius of the circle by $r$, we obtain
$$\begin{aligned} |DF| &= |DQ| - |FQ| = |DO| - |QO| - |FQ| = \frac{1}{2}|PO| - |QO| - \frac{1}{3}|PQ| \\ &= \frac{1}{2}(|PQ| + |QO|) - |QO| - \frac{1}{3}|PQ| = \frac{1}{6}|PQ| - \frac{1}{2}|QO|, \end{aligned}$$
$$\begin{aligned} |FG| &= |FQ| - |GQ| = \frac{1}{3}|PQ| - |GQ| = \frac{1}{3}|PQ| - r + |QO|, \\ |FH| &= |FQ| + |QO| + r = \frac{1}{3}|PQ| + |QO| + r. \end{aligned}$$
From this we obtain
$$\begin{aligned} |FG| \cdot |FH| &= \left(\frac{1}{3}|PQ| + |QO| - r\right) \left(\frac{1}{3}|PQ| + |QO| + r\right) \\ &= \left(\frac{1}{3}|PQ| + |QO|\right)^2 - r^2 \\ &= \frac{1}{9}|PQ|^2 + \frac{2}{3}|PQ| \cdot |QO| + |QO|^2 - r^2 \\ 2|DF| \cdot |FQ| &= \frac{2}{3}|PQ| \left(\frac{1}{6}|PQ| - \frac{1}{2}|QO|\right) = \frac{1}{9}|PQ|^2 - \frac{1}{3}|PQ| \cdot |QO|. \end{aligned}$$
Using that $\triangle APO$ is a right triangle, we get $|PQ| \cdot |QO| = |AQ|^2$. Pythagoras gives $r^2 - |QO|^2 = |AQ|^2$, hence $r^2 - |QO|^2 = |PQ| \cdot |QO|$. Therefore
$$\frac{2}{3}|PQ| \cdot |QO| + |QO|^2 - r^2 = -\frac{1}{3}|PQ| \cdot |QO|$$
and we arrive at
$$|FG| \cdot |FH| = 2|DF| \cdot |FQ|.$$
Because $2|FQ| = |PF|$ and the power of $F$ with respect to the given circle is $|FG| \cdot |FH| = |FC| \cdot |FA|$, we can rewrite this as
$$\begin{aligned} \frac{|FC| \cdot |FA|}{|DF|} &= \frac{|DF| \cdot |PF|}{|FA|} \quad \text{ i.e. } \\ \frac{|FC|}{|DF|} &= \frac{|PF|}{|FA|}. \end{aligned}$$
This shows that the triangles $DFC$ and $AFP$ are similar and so $\angle CDF = \angle PAF$. As $PA$ is tangent to the circle at $A$ we also have $\angle AEC = \angle PAF$, hence $\angle CDF = \angle AEC$ which shows that $PO$ and $AE$ are parallel.