128
Two chords BA and CD intersect at point P outside the circle. Prove that $\Delta PDA \sim \Delta PBC$
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$\angle PDA + \angle ADC = 180^\circ$ (linear pair) ........ (i) [$\frac{1}{2}$ mark]
Also $\angle ADC + \angle ABC = 180^\circ$ (opposite angles of cyclic quadrilateral) .... (ii) [$\frac{1}{2}$ mark]
From (i) and (ii) $\angle PDA = \angle ABC$ [$1$ mark]
Also $\angle P$ is common therefore $\Delta PDA \sim \Delta PBC$ (By AA similarity criterion) [$\frac{1}{2}$ mark]
Also $\angle ADC + \angle ABC = 180^\circ$ (opposite angles of cyclic quadrilateral) .... (ii) [$\frac{1}{2}$ mark]
From (i) and (ii) $\angle PDA = \angle ABC$ [$1$ mark]
Also $\angle P$ is common therefore $\Delta PDA \sim \Delta PBC$ (By AA similarity criterion) [$\frac{1}{2}$ mark]