Question

plz prove it
in steps​

Answer 1

Given,

Two circles that touch each other internally at point P in which Chord AB of the larger circle intersect smaller circle at pint C and D.

To Prove: ∠ CPA = ∠DPB

Construction: Draw a tangent TS at P to the circles.

Since TPS is the tangent, PD is the chord.

∴, ∠PAB=∠BPS ..........(1) [Angle in the alternate segment are always equal]

Similarly, ∠PCD = ∠DPS.......(2)

Now, by subtracting equation (1) from (2),

∠PCD-∠PAB = ∠DPS-∠BPS

But in ΔPAC,

Exterior of ∠PCD = ∠PAB+∠CPA

∴,∠PAB+∠CPA-∠PAB = ∠DPS-∠BPS

⇒∠CPA = ∠DPB

Hence,∠CPA is equal to ∠DPB.