Question

from an external point P two tangents PA and PB are drawn to the circle with Centre O prove that op is a perpendicular bisector of AB

Answer 1

Answer:

Step-by-step explanation:Here is the answer to your query.

 



Let OP intersect AB at C

In ΔPAC and ΔPBC, we have

PA = PB ( Tangent from an external point are equal)

∠APC = ∠BPC ( PA and PB are equally inclined to OP)

and PC = PC  ( Common)

∴ ΔPAC  ΔPBC  (by SAS congurency criteria)

⇒ AC = BC   ......(1)

and ∠ACP = ∠BCP  ......(2)

But ∠ACP + BCP = 180°  .......(3)

From (2) and (3)

∠ACP = ∠BCP =  = 90°

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

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Answer 2

Answer:

Step-by-step explanation: