By the cosine law on $\triangle ADO$ and $\triangle OPC$, we obtain that $AO^{2}=DO^{2}+AD^{2}$ and $OC^{2}=OP^{2}+PC^{2}-2\cdot OP\cdot PC\cdot \cos{\angle OPC}$, hence we need \begin{align*} AD^{2}&=PC^{2}-2\cdot OP\cdot PC\cdot \cos{\angle OPC}\\ AD^{2}&=PC^{2}+2\cdot OP\cdot PC\cdot \cos{\angle DPO}\\ AD^{2}&=PC(PC+2\cdot OP\cdot \cos{\angle DPO})\\ AD^{2}&=PC(PC+DP)\\ BC^{2}&=PC\cdot CD,\end{align*}which is obviously true, since $\triangle BCP\sim\triangle DCB$ as $\angle BDC=\angle DBA = \angle CBP$. Done.

WLOG assume $AB>CD$. Let $\omega$ denote circle passing trough $P,D$ and tangent to $AD$. Note that: $$ \angle DAB = \angle DBC = \angle CBP$$Thus $CB$ is tangent to $\odot(BPD)$. This implies the following: $$ CB^2 = CP \cdot CD = AD^2$$Since $AD$ is tangent to $\omega$, we deduce that points $A, C$ have equal powers with respect to $\omega$. Therefore: \begin{align*} Pow_{\omega}(A) = Pow_{\omega}(C) \\ AO^2 -r^2 = OC^2 -r^2 \\ AO=OC \end{align*}as desired.

Special case of Sharygin CR 2017 P10