Figure 32

As $OQ$ is the perpendicular bisector of $AD$, one has $\angle QAD = \angle ADQ = \angle BDC = \angle CBD$ (Fig. 32). Therefore $AQ \parallel BC$, whence $\angle QAC = 180^\circ - \angle ACB = 90^\circ$. From the cyclic quadrilateral $APCQ$ one also gets $\angle CPQ = \angle PQA = 90^\circ$, i.e., $ACPQ$ is a rectangle.

As $\angle DOC = 2\angle DAC$, one obtains
$$\begin{aligned} \angle DOC &= 2\angle BAC = 2(90^\circ - \angle CBA) = 180^\circ - 2\angle CBA = \\ &= 180^\circ - \angle CBD - \angle BDC = \angle DCB, \end{aligned}$$
which implies that isosceles triangles $BDC$ and $DCO$ are similar. Thus $\angle BDC = \angle DCO$, i.e., $OC \parallel AB$, whence $\angle QOC = 90^\circ = \angle QAC$. So $O$ lies on the circle determined by $A$, $C$, $P$, $Q$. Therefore
$$\angle ACQ = \angle AOQ = \frac{1}{2}\angle AOD = \frac{1}{2}\angle AOP = \frac{1}{2}\angle ACP.$$
Consequently, the diagonal of the rectangle $ACPQ$ bisects the angle of the rectangle, whence $ACPQ$ is a square.