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प्रश्न
In the figure; PA is a tangent to the circle, PBC is secant and AD bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that:
`∠CAD = 1/2 (∠PBA - ∠PAB)`
उत्तर
i. PA is the tangent and AB is a chord
∴ ∠PAB = ∠C ...(i) (Angles in the alternate segment)
AD is the bisector of ∠BAC
∴ ∠1 = ∠2 ...(ii)
In ΔADC,
Ext. ∠ADP = ∠C + ∠1
`=>` Ext ∠ADP = ∠PAB + ∠2 = ∠PAD
Therefore, ΔPAD is an isosceles triangle.
ii. In ΔABC,
Ext. ∠PBA = ∠C + ∠BAC
∴ ∠BAC = ∠PBA – ∠C
`=>` ∠1 + ∠2 = ∠PBA – ∠PAB ...(From (i) part)
`=>` 2∠1 = ∠PBA – ∠PAB
`=> ∠1 = 1/2 (∠PBA - ∠PAB)`
`=> ∠CAD = 1/2 (∠PBA - ∠PAB)`
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