Question

Two circles with centres P and Q intersect at A and B respectively . C lies on the circle with centre at Q as shown in figure. CD is a tangent at A to the circle with centre Q .. Prove that angle cap = angle pab​

Answer 1

Answer:

Join AB and let XY be the tangent at P. Then by alternate segment theorem,

∠APX=∠ABP ……………(i)

Next, ABCD is a cyclic quadrilateral, therefore, by the theorem sum of the opposite angles of a quadrilateral is 180^{\circ}

∠ABD+∠ACD=180

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Also, ∠ABD=∠ABP=180

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(Linear Pair)

∴∠ACD=∠ABP ...........(ii)

From (i) and (ii),

∠ACD=∠APX

∴XY∥CD (Since alternate angles are equal).

solution