Two circles intersect at two points $B$ and $C$. Through $\mathrm{B}$, two line segments $\mathrm{ABD}$ and $\mathrm{PBQ}$ are drawn to intersect the circles at $A, D$ and $P$, $Q$ respectively (see in figure below). Prove that $\angle \mathrm{ACP}=\angle \mathrm{QCD}$."

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Given:

Two circles intersect at two points $B$ and $C$. Through $\mathrm{B}$, two line segments $\mathrm{ABD}$ and $\mathrm{PBQ}$ are drawn to intersect the circles at $A, D$ and $P$, $Q$ respectively.

To do:

We have to prove that $\angle \mathrm{ACP}=\angle \mathrm{QCD}$.

Solution:

We know that,

Angles in the same segment are equal.

In the larger circle,

$\angle ACP =\angle ABP$...…(i)    (Angles in the same segment)

In the smaller circle,

$\angle QCD = \angle QBD$...…(ii)                {Angles in the same segment)

$\angle ABP = \angle QBD$...…(iii)         (Vertically opposite angles)

From (i), (ii) and (iii), we get,

$\angle ACP = \angle QCD$.

Hence proved.

Updated on 10-Oct-2022 13:46:46