Let O is the centre of the circle.

Let C is the mid point of the minor arc of AB and PQ is the tangent to the circle through the point C.

Since C is the mid-point of the arc AB

=> minor arc AB = minor arc BC

=> AC = BC

This shows that ΔABC is an isosceles triangle.

So, the perpendicular bisector of the side AB of ΔABC passes through the vertex C

Again since AB is a chord to the circle, So the perpendicular bisector of AB passes thorughpoint O and C

So, AB ⊥ OC

Now, PQ is the tangent to the circle through the point C on the circle.

So, PQ ⊥ OC   {since tangent to a circle is perpendicular to its radius through the point of contact}

Now, The chord AB and tangent PQ of the circle are perpendicular to the same line segment.

So, AB is parallel to PQ

Hence, the tangent drawn at the end of the mid-point of an Arc of a circle is parallel to the chord going the end point of the

arc.