Prove that the tangent drawn at the mid-point of an arc of a circle is parallel to the chord joining the end points of the arc.

#### Solution 1

In the given figure, C is the mid point of the minor arc AB of the circle with centre O.

PQ is the tangent to the given circle through point C.

To Prove: Tangent drawn at the mid point of the arc

i.e AB || PQ.

Proof: C is the mid point of the minor arc AB

⇒ minor arc AC = minor arc BC

⇒ AC = BC

Thus, ∆ABC is an iscosceles triangle.

Therefore, the perpendicular bisector of side AB of ∆ABC passes through the vertex C.

We know that the perpendicular bisector of a chord passes through the centre of the circle.

Since AB is a chord of the circle so, the perpendicular bisector of AB passes through the centre O.

Thus, it is clear that the perpendicular bisector of AB passes through the points O and C.

Therefore,

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

Therefore,

#### Solution 2

Let us draw a circle in which AMB is an arc and M is the mid-point of the arc AMB.

Joined AM and MB. Also TT' is a tangent at point M on the circle.

**To Prove:** AB || TT'

**Proof:** As M is the mid point of Arc AMB

Arc AM = Arc MB

AM = MB .......[As equal chords cuts equal arcs]

∠ABM = ∠BAM .......[Angles opposite to equal sides are equal] [1]

Now, ∠BMT' = ∠BAM ......[Angle between tangent and the chord equals angle made by the chord in alternate segment] [2]

From [1] and [2]

∠ABM = ∠BMT'

So, AB || TT' [two lines are parallel if the interior alternate angles are equal]

Hence Proved!