AB is tangent to the circle through point C.
We have to show the tangent drawn at the mid-point of the arc PQ of a circle is parallel
to the chord joining the end points of the arc PQ.
We will show PQ ║AB.
It is given that C is the midpoint point of the arc PQ.
So, arc PC = arc CQ.
⇒PC = CQ
This shows that ΔPQC is an isosceles triangle.
Thus, the perpendicular bisector of the side PQ of ΔPQC passes through vertex C.
The perpendicular bisector of a chord passes through the centre of the circle.
So the perpendicular bisector of PQ passes through the centre O of the circle.
Thus the perpendicular bisector of PQ passes through the points O and C.
AB is the tangent to the circle through the point C on the circle.
⇒AB ⊥ OC
The chord PQ and the tangent PQ of the circle are perpendicular to the same line OC.
∴PQ || AB.