Two circles touch internally at point P and a chord AB of the circle of larger radius intersects the other circle in C and D. Which of the following holds good?

∠ C P A = ∠ D P B

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2 ∠ C P A = ∠ C P D

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A P X = ∠ A D P

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∠ B P Y = ∠ C P D + ∠ C P A

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Solution

The correct option is A $∠ C P A = ∠ D P B$
Angle XPA = angle ABP = x
Angle CPX = angle CDP = x + y
Angle CDP is exterior angle of triangle PDB
So angle CDP = DBP + DPB
X + y = x + DPB
DPB = y
So angle CPA = DPB

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Two circles touch internally at point P and a chord AB of the circle of larger radius intersects the other circle in C and D. Which of the following holds good?

The correct option is A ∠CPA=∠DPBAngle XPA = angle ABP = x Angle CPX = angle CDP = x + y Angle CDP is exterior angle of triangle PDB So angle CDP = DBP + DPB X + y = x + DPB DPB = y So angle CPA = DPB

The correct option is A $∠ C P A = ∠ D P B$
Angle XPA = angle ABP = x
Angle CPX = angle CDP = x + y
Angle CDP is exterior angle of triangle PDB
So angle CDP = DBP + DPB
X + y = x + DPB
DPB = y
So angle CPA = DPB