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)