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In the adjoining figure, ABCD is a parallelogram and E is mid-point of AD. DL || EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF.

In the figure, ABCD is a parallelogram and E is mid-point of AD. DL || EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF. Mid-point Theorem, ML Aggarwal Understanding Mathematics Solutions ICSE Class 9.

Mid-point Theorem

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Answer

Given, DL || EB.

Since, DL || BE we can say that,

⇒ BE || DF

In △AFD,

E is midpoint of AD and BE is parallel to DF,

∴ B is midpoint of AF (By converse of midpoint theorem).

In BEDL,

LD || BE and BL || DE

∴ BEDL is a parallelogram.

Since, BEDL is a parallelogram opposite sides are equal.

Let LD = BE = x.

E is midpoint of AD and B is the midpoint of AF

By midpoint theorem,

BE = 12\dfrac{1}{2}FD

FD = 2BE = 2x.

LF = FD - LD = 2x - x = x.

Since, LF = BE = x.

Hence, proved that B is midpoint of AF and EB = LF.

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