Question

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

Answer 1

In triangle ABE and AFD,
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, which makes B the midpoint on 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),
=> LF = BE
Thus proved.