Question

Two circles touch Each Other internally at a point P and chord AB of the greater circle touches the smaller circle at C and D prove that angle ABC is equal to Angle BPD

Answer 1

Prove that ∠CPA = ∠DPB

Explanation:

The given question is mistake. The correct question is:

Two circles touch Each Other internally at a point P and chord AB of the greater circle touches the smaller circle at C and D prove that angle CPA is equal to Angle DPB.

The reference image to the answer is attached below.

Draw a tangent TS at P to the given circles.

Since TPS is a tangent and PD is a chord.

Angeles in alternate segment are equal.

∴ ∠PAB = ∠BPS --------- (1)

Similarly,

∴ ∠PCD = ∠DPS --------- (2)

Subtract equation (1) from equation (2), we get

∠PCD – ∠PAB = ∠DPS – ∠BPS --------- (3)

But in angle PAC,

Ext. ∠PCD = ∠PAB + ∠CPA

Substitute this in equation (3),

∠PAB + ∠CPA – ∠PAB = ∠DPS – ∠BPS

∠CPA = ∠DPS – ∠BPS

we know that ∠DPB = ∠DPS – ∠BPS.

⇒ ∠CPA = ∠DPB

Hence proved.

To learn more...

1. In the figure, two circles touch internally at point P.

Chord AB of the larger circle intersects the smaller

circle in C.  Prove : CPA congrats to angle DPB

https://brainly.in/question/13564757

2. If two circles are internally touching at point P and a line intersect those two line in points A,B,C,D respectively then prove that angle APB is equals to angle CPD

https://brainly.in/question/6892209