Prove that the line segment joining the mid-points of a pair of opposite sides of a parallelogram divides it into two equal parallelograms
Answers
Answer:
In ||gm ABCD, E is the mid-point of AB and F is the mid-point of DC.
Also AB|| DC.
AB = DC and AB || DC
∴ (1/2)AB = (1/2)DC and AE || DF (Since E and F mid point of AB and DC)
∴ AE = (1/2)AB and DF = (1/2)DC
∴ AE = DF and AE || DF
∴Quadrilateral AEFD is a parallelogram
Similarly, Quadrilateral EBCF is a parallelogram.
Now parallelogram AEFD and EBCF are on equal bases DF = FC and between two parallels AB and DC
∴ ar(||gm AEFD) = ar(||gm EBCF)
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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.