From an external point P, two tangents PA and PB are drawn to a circle
Given: PA and PB are tangents to a circle with centre O
Let AB and OP intersect at R.
In ∆APR and ∆BPR
PA = PB (tangents from an external point to a circle are equal in length)
∠APR = ∠ BPR (tangents from an external point are equally inclined to the point joining the centre of the circle)
PR = PR (Common)
∆APR ≅ ∆BPR (By SAS congruency criterion).
⇒ AR = BR ............ (1)
and ∠ARP = ∠BRP
But ∠ARP + ∠BRP = 180°
from (1) and (2) we can conclude that OP is the perpendicular bisector of AB