In the given figure, m and M is the midpoint of AB. Prove that M is also the midpoint of any line segment CD having its end points at and m respectively.

Here it is given that l m ie. AC ||DB.


Also given that AM = MB


Now in ∆AMC and ∆BMD,


CAM = DBM … Alternate angles


AM = MB


AMC = BMD … vertically opposite angles


Hence, ∆AMC BMD … by ASA property of congruency


CM = MD …cpct


Hence proved that M is also the midpoint of any line segment CD having its end points at and m respectively.


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