# In fig ABCD is a parallelogram and E is the mid point of AD.DL is parallel to BE meets Ab is producer at F. Prove that B os the mid point of AF and EB=LF

1
is L the mid point of BC?

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is L the mid point of BC?

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EB is parallel to DF

=> Angle EAB = Angle DAF (common)

=> Angle ABE = Angle AFD (corresponding angles)

Therefore triangle ABE is similar to triangle AFD, by Angle-Angle test for similarity.

=> AB/AF = AE/AD (CPST)

But, E is midpoint of AD

Therefore AE = AD/2

=> AE/AD = 1/2

Therefore, from the CPST equation,

=> AB/AF = 1/2

=> AB = AF/2

Therefore AB is half of AF,

Similarly,

=> AE/AD = BE/FD (CPST)

=> 1/2 = BE/FD

=> BE = FD/2 ...(1)

AB = CD (opposite sides of parallelogram)

and

AB = FB (equal halves of AF)

Therefore CD = FB

Now, proceed to prove triangles BFL and CDL congruent by Angle-Side-Angle test for congruency.

Therefore, DL = LF (CPCT)

This makes L midpoint of FD

=> LF = FD/2

From (1),

=>

Thus proved.