PD tangent circle o at point d

∴OD⊥PD

∵C is the midpoint of the semicircle ABC.

∴OC⊥AB

OC = OD

∴∠OCE=∠ODE

∠∠OCE+∠OEC = 90°

∠ODE+∠PDE=90

∴∠OEC=∠PDE

Similarly ∠OEC=∠DEP

∴∠PDE=∠DEP

∴PD=PE

(2) Connecting AD and BD

∫AB is the diameter of circle O, and D is a point on the circle.

∴AD⊥BD

∴∠DAB+∠DBA=90

∵PD⊥OD

∴∠BDP+∠ODB=90

OD=OB

∴∠DBA=∠ODB

∴∠DAB=∠BDP

And ∠P is the common angle of ⊿BDP and ⊿DAP.

Available ⊿BDP∽⊿DAP

∴PD/PA=PB/PD

Namely PD? =PA×PB

As can be seen from the above questions, PD=PE.

∴PE？ =PA×PB

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