(ii)
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
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Answer:
proved
Step-by-step explanation:
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Answered by
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∠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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