In the figure,two chords $AB$ and $CD$ of a circle intersect each other at point $P$ (when produced) outside the circle. Prove that $\Delta PAC \sim \Delta PDB$.
(N/A) Consider $\Delta PAC$ and $\Delta PDB$.
$1$. $\angle P = \angle P$ (Common angle).
$2$. In a cyclic quadrilateral $ABDC$,the exterior angle is equal to the interior opposite angle. Therefore,$\angle PAC = \angle PDB$ (Since $\angle PAC + \angle CAB = 180^{\circ}$ and $\angle PDB + \angle CAB = 180^{\circ}$).
$3$. By the $AA$ (Angle-Angle) similarity criterion,$\Delta PAC \sim \Delta PDB$.
$1$. $\angle P = \angle P$ (Common angle).
$2$. In a cyclic quadrilateral $ABDC$,the exterior angle is equal to the interior opposite angle. Therefore,$\angle PAC = \angle PDB$ (Since $\angle PAC + \angle CAB = 180^{\circ}$ and $\angle PDB + \angle CAB = 180^{\circ}$).
$3$. By the $AA$ (Angle-Angle) similarity criterion,$\Delta PAC \sim \Delta PDB$.
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