the line segment joining the mid point of any two sides of a triangle is parallelogram to the third sides and equal to half of it
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Let us consider ABCD be a parallelogram in which E and F are mid-points of AB and CD. Join EF.
To prove: ar (|| AEFD) = ar (|| EBCF)
Let us construct DG ⊥ AG and let DG = h where, h is the altitude on side AB.
Proof:
ar (|| ABCD) = AB × h
ar (|| AEFD) = AE × h
= ½ AB × h ….. (1) [Since, E is the mid-point of AB]
ar (|| EBCF) = EF × h
= ½ AB × h …… (2) [Since, E is the mid-point of AB]
From (1) and (2)
ar (|| ABFD) = ar (|| EBCF)
Hence proved.
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