Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that : $∠ C P A = ∠ D P B$ .

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Solution

Given - Two circles touch each other internally at P. A chord AB of bigger circle intersects the smaller circle at C and D. AP, BP , CP and DP are joined.

To Prove - ∠CPA =∠DPB

Construction - Draw a tangent TS at P to the circles given.

Proof -

As, TPS is the tangent , PD is the chord.

∴ ∠PAB = ∠BPS..........(i) [Angles in alt. segment]

Similarly, we can prove that ∠PCD = ∠DPS ....(ii)

subtract (i) from (ii),we get,

∠PCD- ∠PAB = ∠DPS - ∠BPS ......(iii)

But in $△ P A C$

∠PCD = ∠PAB + ∠CPA .....(iv) (external angle equals sum of interior opposite angles)

therefore, from (iii) and (iv)

∠PAB + ∠CPA - ∠PAB = ∠DPS- ∠BPS

∠CPA = ∠DPB

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Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that : ∠ CPA = ∠DPB.

Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Prove that : $∠ C P A = ∠ D P B$ .