$\rm \angle PAO+\angle OAQ=2\cdot 90^{\circ}=180^{\circ}$ so the points $\rm P,A,Q $ are collinear. Also $\rm \angle OPT+\angle OQT=2\cdot 90^{\circ}=180^{\circ}$ so the points $\rm T,P,O,Q$ are concyclic . $\rm \angle POD=\angle OAQ=90^{\circ}-\angle AQO=\angle TQP=\angle TOP$ q.e.d

We will prove this theorem for any non-degenerate triangle $ABC$. All angles in this solution are directed. $$\measuredangle QCO=90^\circ\implies\text{QC is tangent to ABC circumcircle}$$and $$\measuredangle PAO=90^\circ\implies\text{PA is tangent to ABC circumcircle.}$$By tangent-chord angle theorem $$\measuredangle POD= \measuredangle PAB+ \measuredangle BAD= \measuredangle ACB+ \measuredangle BAC= \measuredangle ABC,\ \measuredangle ACQ= \measuredangle ABC.$$Observe$$\measuredangle TPO=90^\circ= \measuredangle TQO\implies\text{points O, P, Q, T are concyclic}\implies \measuredangle POT= \measuredangle PQT= \measuredangle AQT.$$Finally $QT$ being tangent to $c_2$ implies $$ \measuredangle AQT= \measuredangle AOQ= \measuredangle ACQ= \measuredangle ABC.$$Therefore $$ \measuredangle POT= \measuredangle ABC= \measuredangle POD.$$QED