Mathematics
In the given figure, ABCD is a parallelogram and E is the mid-point of BC. If DE and AB produced meet at F, prove that AF = 2AB.
Answer
Given,
ABCD is a parallelogram.
E is the midpoint of BC.
DE meets AB produced at F.
AB ∥ DC
AB = DC (opposite sides of a parallelogram are equal)
Since E is the midpoint of BC :
BE = EC
In △DEC and △FEB:
EC = EB (E is mid-point of BC)
∠DEC = ∠FEB (Vertically opposite angles are equal)
∠DCE = ∠FBE (Alternate interior angles are equal)
△DEC ≅ △FEB (By A.S.A. axiom)
DC = BF [By C.P.C.T.C.]
DC = AB
∴ BF = AB
From figure,
AF = AB + BF
AF = AB + AB
AF = 2AB.
Hence, proved that AF = 2AB.
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