In Fig. 14.36, ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F, prove that AF=2AB.
Answers
Given : ABCD is a parallelogram and E is the mid-point of side BC. If DE and AB when produced meet at F.
AB = CD and AD = BC
[In a parallelogram opposite sides are equal]
BE = EC
[E is the mid point of side BC]
To prove :
AF = 2AB
PROOF :
In ΔDEC and ΔFEB
BE = EC (E is the mid point)
∠DEC = ∠BEF (vertically opposite angles)
∠DCE = ∠FBE
[DC || AF and BC is transversal , so alternate interior angles]
Therefore , ΔDEC ≅ ΔFEB
[By ASA congruence criterion]
So, DC = FB [By CPCT] …………(1)
Also ,we have DC = AB (given) ……….(2)
From eq (1) & (2) , we get :
AB = BF
Adding AB to both sides :
AB + AB = BF + AB
2AB = AF
Hence proved….
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