Question

From an external point P tow tangents PA and PB are drawn to the drawn to the circle with centre O. prove that OP is the perpendicular bisector of AB.​

Answer 1

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Answer 2

Let us assume that AB and OP intersect each other at point M.

In ∆APM and ∆BPM

PA = PB (length of tangents drawn from an external point to a circle are equal to each other)

PM = PM (Common)

∠APM = ∠ BPM (tangents drawn from an external point to a circle are equally inclined to the segment joining the centre of the circle of that point)

By SAS

∆APM ~ ∆BPM

By CPCT

AM = BM  

Also, ∠AMP = ∠BMP

As AB is a line (so makes an angle of 180°)

→ ∠AMP + ∠BMP = 180°

→ ∠AMP + ∠AMP = 180° (As ∠AMP = ∠BMP )

→ 2∠AMP = 180°

→ ∠AMP = 180°/2

→ ∠AMP = 90°

Therefore,

∠AMP = ∠BMP = 90°

This proves that,

OP is the perpendicular bisector of AB.