In the figure,two chords $AB$ and $CD$ of a circle are produced to intersect each other at point $P$ outside the circle. Prove that $PA \cdot PB = PC \cdot PD$.

(N/A) To prove $PA \cdot PB = PC \cdot PD$,consider $\triangle PAD$ and $\triangle PCB$.
$1$. $\angle P = \angle P$ (Common angle).
$2$. $\angle PDA = \angle PBC$ (Exterior angle of a cyclic quadrilateral is equal to the interior opposite angle,or since $\angle PAD + \angle DAB = 180^{\circ}$ and $\angle BCD + \angle DAB = 180^{\circ}$,we have $\angle PAD = \angle BCD$,and thus $\triangle PAD \sim \triangle PCB$ by $AA$ similarity).
Since $\triangle PAD \sim \triangle PCB$,the ratios of their corresponding sides are equal:
$\frac{PA}{PC} = \frac{PD}{PB} = \frac{AD}{CB}$.
Taking the first two parts:
$\frac{PA}{PC} = \frac{PD}{PB}$
By cross-multiplying,we get:
$PA \cdot PB = PC \cdot PD$.