Question
In given figure, two circles intersect at two points $B$ and $C$. Through $B$, two line segments $A B D$ and $P B Q$ are drawn to intersect the circles at $A, D$ and $P, Q$ respectively. Prove that $\angle A C P=\angle Q C D$.
✓
Answer
As angles in the same segment of circle are equal
$\angle ABP = \angle ACP ..........................(i)$
$\angle ABP =\angle QBD$ (Vertically opposite angles)
Also, $\angle QCD =\angle QBD$ (Angles in the same segment)
$\therefore \angle ABP=\angle QCD..........(ii)$
From $(i)$ and $(ii),$ we have
$\angle ACP =\angle QCD$.
$\angle ABP = \angle ACP ..........................(i)$
$\angle ABP =\angle QBD$ (Vertically opposite angles)
Also, $\angle QCD =\angle QBD$ (Angles in the same segment)
$\therefore \angle ABP=\angle QCD..........(ii)$
From $(i)$ and $(ii),$ we have
$\angle ACP =\angle QCD$.
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