In Figure, L || 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.

Given: L || m and M is mid point of segment AB.

As L || m,

∠BAC = ∠ABD (alternate interior angle)

∠AMC = ∠DMB (vertically opposite angle)

In ΔAMC and ΔBMD

∠BAC = ∠ABD (proved)

∠AMC = ∠DMB (proved)

AM = BM (given)

Hence, ΔAMC ≅ΔBMD (by ASA)

⇒ MC = MD (by CPCT)