Question

In the figure above, two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively. Prove that angle ACP= angle QCD​

Answer 1

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$\huge\bold\purple{Answer:-}$

⏩ Since angles in the same segment of a circle are equal.

Therefore,

angle ABP = angle ACP → (i)

But, angle ABP = angle DBQ (Vertically opposite angles)

and, angle DBQ = angle DCQ (Angles in the same segment)

Therefore,

angle ABP = angle DCQ → (ii)

Now, from (i) and (ii), we have

angle ACP = angle DCQ => angle ACP= angle QCD

[Hence proved]

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Answer 2

Answer:

angle ABP = angle QBD. { vert. opp. angles }

angle ACP = angle ABP { .°. angle in the same segment are equal }

angle QCD = angle QBD { .°. angle in the same segment are equal }

Now, see carefully each angle is equal with each other, so we get

angleQBD = angleABP = angleACP = angleQCD.

.°. angle ACP = angle QCD, as required