\(l \parallel m\) 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.
Proof:
\(\triangle AMC \cong \triangle BMD\) [By Congruence rule]
Then, \(MC = MD\) [By ]
This implies that \(M\) is also the midpoint of \(CD\).
Hence, we proved.