Solution:

Since $PQ$ is parallel to $BC$, there is a dilation (centred at $A$) of factor $AB / AP$ that sends triangle $APQ$ to triangle $ABC$. Thus
$$|PAQ| = |ABC| \times \left(\frac{AP}{AB}\right)^{2}.$$
Since $AD$ is parallel to $BC$, triangle $ABC$ has the same height as triangle $BCD$. Triangles $ABC$ and $BCD$ also have a common base, so $|ABC| = |BCD|$. Similarly $|DAQ| = |DAP|$ (because $PQ \parallel AD$ and $AD$ is a common base).
$$\therefore |DAP| = |DAB| \times \frac{AP}{AB}$$
$$\qquad = |BCD| \times \frac{AP}{AB}$$
$$\qquad |DAP|^{2} = |BCD| \times |BCD| \times \left(\frac{AP}{AB}\right)^{2}$$
$$\qquad = |BCD| \times |PAQ| \qquad \mathrm{(diagonal~splits~parallelograms~in~half)}$$
$$\qquad = |BCD| \times |PAQ| \qquad \mathrm{(because~}|PAQ| = |ABC|(AP / AB)^{2})$$
as required.