Question

In the given fig. PA and PB are tangents to the circle with centre O. Prove that OP bisects AB and is perpendicular to it.

Answer 1

GIVEN: PA and PB are tangents to a circle with centre O
Let AB and OP intersect at M.

In ∆APM and ∆BPM
PA = PB            
 [the Length of tangents drawn from an external point to a circle are equal.]
∠APM = ∠ BPM      
 [tangents from an external point are equally inclined to the point joining the centre of the circle]
PM = PM         (Common)
∆APM ≅ ∆BPM     (By SAS criterion of  congruence).
AM = BM   & ∠AMP = ∠BMP  ............ (1)
[CPCT]

But ∠AMP + ∠BMP = 180° [ AB is a straight line]
∠AMP + ∠AMP = 180°
2∠AMP = 180°
∠AMP = 180°/2 = 90°
∠AMP = 90°
∠AMP = ∠BMP = 90°……………….(2)

From eq  (1) and (2) , OP bisects AB and  is perpendicular to AB.

HOPE THIS WILL HELP YOU...