$解：連接OP$
$\because O C=O P$
$\therefore \angle O C P=\angle O P C=(180^{\circ}-\angle O) \div 2=90^{\circ}-\frac {1}{2} \angle O$
$\therefore P F \perp O A$
$\because \angle P F C=90^{\circ}$
$則 \angle O C P=90^{\circ}-\angle F P C$
$\because P F \perp O A$
$\therefore \angle P F C=90^{\circ}$
$則 \angle O C P=90^{\circ}-\angle F P C$
$\therefore \angle F P C=\frac {1}{2} \angle O$
$\because \angle F P C=\angle C P A $
$\therefore \angle O=2 \angle F P C=\angle F P C+\angle C P A=\angle A P F$
$\because \angle A+\angle A P F=90^{\circ}$
$\therefore \angle O+\angle A=90^{\circ}$
$\therefore \angle O P A=180^{\circ}-(\angle O+\angle A)=90^{\circ}$
$\therefore P A 是 \odot O 的切線$
