Question

In the figure, two circles touch internally at point P.
Chord AB of the larger circle intersects the smaller
circle in C.
Prove: angleCPA = angleDPB.​

Answer 1

QR is tangent to the circle

So,

Angle PAB=Angle BPQ - 1

Similarly,

Angle PCD=Angle DPQ -2

Subtract 1 from 2

PCD-PAB=DPQ-BPQ

but AnglePCD=PAB+CPA

PAB+CPA-PAB=DPQ-BPQ

Hence,

Angle CPA=Angle DPB

Answer 2

Answer:-

CD=BD

Explanation:-

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