Question

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.

Answer 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

\[\stackrel\frown{AB} \] of a circle is parallel to the chord joining the end point of the arc \[\stackrel\frown{AB} \]

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,

\[AB \perp OC\]

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

\[PQ \perp OC \left[ \text{Tangent to a circle is perpendicular to its radius through the point of contact} \right]\]

Answer 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!