85
In the given figure, $PT$ is a tangent to the circle centered at $O$. $OC$ is perpendicular to chord $AB$. Prove that $PA \cdot PB = PC^2 - AC^2$.

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$PA \cdot PB = (PC - AC) \cdot (PC + BC)$
$= (PC - AC) \cdot (PC + AC)$
quad $[AC = BC]$
$= PC^2 - AC^2$
$= (PC - AC) \cdot (PC + AC)$
quad $[AC = BC]$
$= PC^2 - AC^2$