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.

(N/A) Let $M$ be the mid-point of an arc $AMB$ and $TMT'$ be the tangent to the circle at $M$.
Join $AB$,$AM$,and $MB$.
Since,$\text{arc } AM = \text{arc } MB$,
$\Rightarrow$ Chord $AM =$ Chord $MB$.
In $\triangle AMB$,$AM = MB$.
$\Rightarrow \angle MAB = \angle MBA$ (angles opposite to equal sides are equal) $\dots(i)$.
Since $TMT'$ is a tangent line at $M$,by the alternate segment theorem:
$\angle AMT = \angle MBA$ (angles in alternate segments are equal).
From $(i)$,$\angle AMT = \angle MAB$.
But $\angle AMT$ and $\angle MAB$ are alternate interior angles formed by the transversal $AM$ intersecting lines $AB$ and $TMT'$.
Since the alternate interior angles are equal,the lines must be parallel.
Therefore,$AB \parallel TMT'$.
Hence,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.