Proof Join the segments $XY$ and $DX$. It follows from the given conditions that $B$, $P$, $D$, $X$ are concyclic, and $C$, $Y$, $Q$, $D$ are concyclic. Then
$$\angle AXM = \angle BXP = \angle BDP \\ = \angle QDC = \angle AYN.$$

And it follows from $\angle AMX = \angle ANY = 90^\circ$ that $\triangle AMC \sim \triangle ANY$, so $\angle MAX = \angle NAY$ and $\frac{AM}{AX} = \frac{AN}{AY}$, and hence $\angle MAN = \angle XAY$.

Combining the two results above, we have $\triangle AMN \sim \triangle AXY$. So we get
$$\begin{aligned} \triangle AMN \sim \triangle ABC &\Leftrightarrow \triangle AXY \sim \triangle ABC \\ &\Leftrightarrow XY \parallel BC \Leftrightarrow \angle DXY = \angle XDB. \end{aligned}$$
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As $A$, $X$, $D$, $Y$ are concyclic, we have $\angle DXY = \angle DAY$.
As $\angle XDB = 90^\circ - \angle ABC$,
$$\angle DXY = \angle XDB \iff \angle DAC = 90^\circ - \angle ABC,$$
which is equivalent to the fact that the line $AD$ passes through the circumcenter of $\triangle ABC$.
Hence, $\triangle AMN \sim \triangle ABC$ if and only if $AD$ passes through the circumcenter of $\triangle ABC$.