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.

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Updated on: 10-Oct-2022

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