Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Then, find the relation between $∠ C P A$ and $∠ D P B$ .

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Solution

Given - Two circles touch each other internally at P. A chord AB of the bigger circle, intersects the smaller circle at C and D, AP, BP, CP and DP are joined.

Construction - Draw a tangent TS at P to the circles given.

$∵ T P S$ is the tangent, PD is the chord.

$∴ ∠ P A B = ∠ B P S$ ....(i) ( 1 mark)

(Angles in alternate segment)

Similarly we can prove that

$∠ P C D = ∠ D P S$ .....(ii) (1 marks)

Subtracting (i) and (ii), we get

$∠ P C D - ∠ P A B = ∠ D P S - ∠ B P S$

But in $Δ P A C$

Ext. $∠ P C D = ∠ P A B + ∠ C P A$

$∴ ∠ P A B + ∠ C P A - ∠ P A B = ∠ D P S - ∠ B P S$

$∠ C P A = ∠ D P B$
so, $∠ C P A$ and $∠ D P B$ are equal (1 mark)

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Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Then, find the relation between ∠CPA and ∠DPB .

Two circles touch each other internally at a point P. A chord AB of the bigger circle intersects the other circle in C and D. Then, find the relation between $∠ C P A$ and $∠ D P B$ .