In the given figure,$l \parallel m$ and $M$ is the mid-point of a line segment $AB$. Show that $M$ is also the mid-point of any line segment $CD,$ having its end points on $l$ and $m,$ respectively.

(N/A) In $\triangle AMC$ and $\triangle BMD$,we have:
$\angle MAC = \angle MBD$ (Alternate interior angles,since $l \parallel m$)
$\angle AMC = \angle BMD$ (Vertically opposite angles)
$AM = BM$ (Given,as $M$ is the mid-point of $AB$)
Therefore,$\triangle AMC \cong \triangle BMD$ (By $ASA$ congruence rule)
Thus,$CM = DM$ (By $CPCT$)
Hence,$M$ is also the mid-point of $CD.$