Question

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.​

Answer 1

Hence proved

∠CPA = ∠DPB

Step-by-step explanation:

In the figure, two circles touch internally at point P. Chord AB of the larger circle intersects the smaller circle at C and D.

To prove: $\angle CPA=\angle DPB$

Construction: Draw a common tangent RS at point P

Proof:

RS is a tangent and ΔAPB is triangle of bigger circle.

$\angle BPS=\angle PAB$       (Angles in alternate segments)

RS is a tangent and ΔCPD is triangle of inner circle.

$\angle DPS=\angle PCD$       (Angles in alternate segments)

Subtract both equation

$\angle DPS - \angle BPS = \angle PCD-\angle PAB$

$\text{But } \angle PCD=\angle PAB+\angle CPA$    

Exterior angle theorem of triangle APC.

$\angle DPB=\angle PAB+\angle CPA-\angle PAB$    

Therefore,

$\angle DPB=\angle CPA$    

hence proved

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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