Question

(11)
In the figure, two circles touch internally at point P.
Chord AB of the larger circle intersects the smaller
circle in C.
Prove : CPA = DPB.
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Answer 1

∠CPA is equal to ∠DPB.

Step-by-step explanation:

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.

Answer 2

Answer:∠CPA is equal to ∠DPB.

Step-by-step explanation:

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.

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