from an external point p, two tangent pa and pb are drawn to the circle with centre o. if c is midpoint of cord ab prove that pc passes through the centre o of the circle

The figure below shows two tangents PA and PB drawn from a point P to a circle with centre O;

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)
So, $\mathrm{PAC}\cong \mathrm{PBC}$  {By SAS congruency}
⇒ AC = BC   ......(1)
and ∠ACP = ∠BCP  ......(2)
But ∠ACP + BCP = 180°  .......(3)

From (2) and (3)
∠ACP = ∠BCP = $\frac{180°}{2} = 90°$
Hence, from (1) and (2), we can conclude that PC is the perpendicular bisector  of AB.
And we know that the line drawn from the centre to the mid point of the chord is perpendicular to the chord.
This means OC is the perpendicular bisector of the chord.
So, PC and OC are the perpendicular bisector of AB. Therefore PC passes through the centre O.