Question

FROM AN EXTERNL PONT P TWO TANGENT PA PB ARE DRAWN WITH ACORCLE TO A CENTE O SHOW THAT OP IS PERPENDICULAR BISECTOR OF AN

Answer 1

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°

"hay second step on pic"

from (1) and (2) we can conclude that OP is the perpendicular bisector of AB