In the figure,two chords $AB$ and $CD$ intersect each other at the point $P$. Prove that $AP \cdot PB = CP \cdot DP$.

(N/A) To prove $AP \cdot PB = CP \cdot DP$,we consider the triangles $\Delta APC$ and $\Delta DPB$.
$1$. $\angle APC = \angle DPB$ (Vertically opposite angles).
$2$. $\angle CAP = \angle BDP$ (Angles in the same segment of a circle are equal).
By the $AA$ (Angle-Angle) similarity criterion,$\Delta APC \sim \Delta DPB$.
Since the corresponding sides of similar triangles are proportional,we have:
$\frac{AP}{DP} = \frac{PC}{PB} = \frac{AC}{DB}$
Taking the first two ratios:
$\frac{AP}{DP} = \frac{PC}{PB}$
By cross-multiplying,we get:
$AP \cdot PB = PC \cdot DP$ (or $AP \cdot PB = CP \cdot DP$).