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Grade: 10
        
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

Answers : (1)

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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