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)


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