Q. 114.9( 7 Votes )
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 atﺎ and m respectively.
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