In the given figure , BA and BA are the tangents to the circle. Prove that OD is the perpendicular bisector of AC, where O is the centre of the circle.

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Solution

In $△$ OBC and $△$ OBA, we have:

$∠$ OCB = $∠$ OAB = 90 $°$ (Tangent is perpendicular to the radius)
OC = OA (Radius of the larger circle)
BC = BA (Tangents drawn from the external point are equal)
∴ $△$ OBC $≅$ $△$ OBA (By SAS congruency)
i.e. $∠$ ABO = $∠$ CBO (CPCT)

In $△$ DBC and $△$ DBA, we have:
$∠$ ABO = $∠$ CBO (Proven above)
BD = BD (Common)
BC = BA (Tangents drawn from the external point are equal)
$△$ DBC $≅$ $△$ DBA (By SAS congruency)
i.e., CD = DA (CPCT)
So, D is the midpoint of AC. ....(i)

The line joining the centre and the midpoint of the chord is perpendicular to the chord.
Hence, $∠$ BDC = 90 $°$
∴ OD is the perpendicular of AC ....(ii)
From (i) and (ii),
OD is the perpendicular bisector of AC.

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