Prove that a diameter $AB$ of a circle bisects all those chords which are parallel to the tangent at the point $A$.

(N/A) Given: $AB$ is a diameter of the circle with center $O$.
Let $MAN$ be the tangent to the circle at point $A$.
Let $CD$ be any chord of the circle that is parallel to the tangent $MAN$.
Since $OA$ is the radius at the point of contact $A$,we have $OA \perp MAN$.
Since $CD \parallel MAN$ and $OA \perp MAN$,it follows that $OA \perp CD$.
Let the diameter $AB$ intersect the chord $CD$ at point $E$.
Thus,$OE \perp CD$.
According to the theorem,the perpendicular drawn from the center of a circle to a chord bisects the chord.
Therefore,$CE = ED$.
Since $CD$ was an arbitrary chord parallel to the tangent at $A$,the diameter $AB$ bisects all such chords.