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`        Two circles internally touch at P. A line intersects the bigger one at A and D and the smaller one at B and C. Prove that angle APB = CPD.`
3 months ago

Arun
23806 Points
```							Given :- circles touch each other internally at P. A chord AB of bigger circle intersects the smaller circle at C and D. AP, BP , CP and DP are joined. To Prove - ∠CPA =∠DPB Construction - Draw a tangent TS at P to the circles given. Proof -  As, TPS is the tangent , PD is the chord. ∴ ∠PAB = ∠BPS..........(i) [Angles in alt. segment] Similarly, we can prove that ∠PCD = ∠DPS ....(ii) subtract (i) from (ii),we get,  ∠PCD- ∠PAB = ∠DPS - ∠BPS But in tr. PAC, ext. ∠PCD = ∠PAB + ∠CPA therefore,   ∠PAB + ∠CPA - ∠PAB = ∠DPS- ∠BPS  ∠CPA = ∠DPB  hence the result.
```
3 months ago
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### Course Features

• 731 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions