[asy][asy] size(10cm); pathpen=black; pointpen=black; pair O=origin, A=(2,0), B=dir(60), C=dir(300), P=dir(20); D(unitcircle); label("$\Gamma$",dir(160),dir(160)); D(MP("B",B,dir(60))--MP("C",C,dir(300))--MP("A",A,E)--B--MP("O",(0,0),W)--C--MP("P",P,dir(20))--B); D(P--MP("F",foot(P,A,B),NE),linetype("0 2")); D(P--MP("D",foot(P,B,C),W),linetype("0 2")); D(P--MP("E",foot(P,C,A),SE),linetype("0 2")); [/asy][/asy] Note that $ABOC$ is cyclic because $\angle OBA = \angle OCA = \dfrac{\pi}{2}$. So, $\angle BOC = \pi-\angle A$ and $\angle OCB = \dfrac{\angle A}2.$ Also, since $\angle BOC = \pi-\angle A,$ angle chasing in circle $\Gamma$ gives $\angle BPC = \dfrac{\pi}2+\dfrac{\angle A}{2}.$ Let $\angle DPC = \alpha.$ Then $\angle DCP = \dfrac{\pi}2-\dfrac{\alpha}2.$ Since the angles in $PBOC$ add up to $2\pi$, \[\angle BOC+\angle OCP+\angle CPB+\angle PBO=2\pi\] \[\pi-\cancel{\angle A}+\cancel{\dfrac{\angle A}2}+\dfrac{\pi}2-\alpha+\dfrac{\pi}2+\cancel{\dfrac{\angle A}2}+\angle PBO = 2\pi\] \[\angle PBO = \alpha=\angle CPD.\] Similarly, $\angle PCO = \angle BPD.$ Since $\triangle PFB \sim \triangle PDC$ and $\triangle PEC \sim \triangle PDB,$ we have that $FPDB \sim DPEC.$ Thus, \[\dfrac{PF}{PD}=\dfrac{PD}{PE}\implies PD^2=PF\cdot PE\] as desired. $\Box$