Question

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

Answer 1

(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.