From an external point P, two tangents PA and PB are drawn to a circle

with centre O as shown in figure. Show that OP is the perpendicular
bisector of AB.

 

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

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