(N/A) Given: $AB$ and $CD$ are two chords of a circle with centre $O$. The mid-points of $AB$ and $CD$ are $L$ and $M$ respectively.
To prove: $AB \parallel CD$
Proof: Since $L$ is the mid-point of chord $AB$,therefore $OL \perp AB$,or $\angle ALO = 90^{\circ}$.
[Because the line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord]
Similarly,$\angle CMO = 90^{\circ}$.
Therefore,$\angle ALO = \angle CMO = 90^{\circ}$.
Since these are corresponding angles formed by the transversal $LM$ intersecting lines $AB$ and $CD$,and they are equal,the lines must be parallel.
So,$AB \parallel CD$.
Hence,proved.