Question

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

$QUESTION$

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.

$ANSWER$ (SHORT ANSWER)

⇒Chords AP and DQ are joined.

⇒For chord AP,

⇒∠PBA=∠ACP        ...Angles in the same segment ---(i)

⇒For chord DQ,

⇒∠DBQ=∠QCD        ...Angles in same segment --- (ii)

⇒ABD and PBQ are line segments intersecting at B.

⇒∠PBA=∠DBQ        ...Vertically opposite angles --- (iii)

By the equations (i), (ii) and (iii),

∠ACP=∠QCD

$SOLUTION$

Given: Two circles intersect at two points B & C. Through B, two line segments ABD & PBQ are drawn which intersect the Circles at A, D, P & Q.

To Prove:

∠ACP = ∠QCD

Proof:

In circle I,

For chord AP,

∠PBA = ∠ACP (Angles in the same segment are equal) — (i)

In circle II,

For chord DQ,

∠DBQ = ∠QCD (Angles in same segment) — (ii)

ABD and PBQ are line segments intersecting at B.

∠PBA = ∠DBQ (Vertically opposite angles) —iii

From the equations (i), (ii) and (iii),

∠ACP = ∠QCD

HENCE THE ANSWER IS ∠ACP = ∠QCD

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