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Two circles intersect at A and B. From a point P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.

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Let XY be the tangent at P.

TPT: CD is || to XY.

Construction: Join AB.

ABCD is a cyclic quadilateral.

∠BAC + ∠BDC= 180° … (1)

∠BDC = 180° – ∠BAC … (2)

Equating (1) and (2)

We get ∠BDC = ∠PAB

Similarly we get ∠PBA = ∠ACD

As XY is tangent to the circle at ‘P’

∠BPY = ∠PAB (by alternate segment there)

∴ ∠PAB = ∠PDC

∠BPY = ∠PDC

XY is parallel of CD.

Hence proved.

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