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प्रश्न
In the figure, PM is a tangent to the circle and PA = AM. Prove that:
(i) Δ PMB is isosceles
(ii) PA x PB = MB2
उत्तर
(i) In Δ PAM,
∠ APM = ∠ AMP ....(i)
PA = AM ...(Given)
by alternate segment property of tangent
∠ ABM = ∠ AMP
∠ APM = ∠ ABM ...(from (i) and (ii))
PM = MB
i.e., ΔPMB is an isosceles ...(proved)
(ii) By rectangle property of tangent and chord,
PM2 = PA x PB
MB2 = PA x PB
Hence proved.
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