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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. Prove that: ∠CPA = ∠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. Prove that: ∠CPA = ∠DPB
Class 10 | Exercise 18B | ICSE | Maths | Selina | Tangents and Intersecting Chords
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. Prove that: ∠CPA = ∠DPB

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(B) - 11

Solution:

Selina Solutions Concise Class 10 Maths Chapter 18 ex. 18(B) - 12

Let’s draw a tangent \[TS\text{ }at\text{ }P\]to the circles given.

As \[TPS\]is the tangent and \[PD\]is the chord, we have

\[\angle PAB\text{ }=\angle BPS\text{ }\ldots .\text{ }\left( i \right)\] [Angles in alternate segment]

Similarly,

\[\angle PCD\text{ }=\angle DPS\text{ }\ldots .\text{ }\left( ii \right)\]

Now, subtracting (i) from (ii) we have

\[\angle PCD\text{ }-\angle PAB\text{ }=\angle DPS\text{ }-\angle BPS\]

But in \[\vartriangle PAC\]

Ext. \[\angle PCD\text{ }=\angle PAB\text{ }+\angle CPA\]

Or,

\[\angle PAB\text{ }+\angle CPA\text{ }\angle PAB\]

\[=\angle DPS\text{ }-\angle BPS\]

Thus,

\[\angle CPA\text{ }=\angle DPB\]

i

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