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In adjoining figure ,ABCD is a parallelogram and E is midpoint of AD .A line through D, drawn parallel to EB, meets AB produced at F and BC at L . Prove that 1.AF=2CD and DF=2DL.

Posted by Suman Kumar 6 years, 2 months ago

CBSE > Class 09 > Mathematics

2 answers

Sia ? 6 years, 2 months ago

It is given that, EB {tex}\parallel{/tex} DL and ED {tex}\parallel{/tex} BL. Therefore, EBLD is a parallelogram.
{tex}\triangle{/tex}DCL {tex}\cong{/tex} {tex}\triangle{/tex}FBL
{tex}\Rightarrow{/tex} DC = BF and DL = FL
Now, BE = DC = AB
{tex}\Rightarrow{/tex} 2AB = 2DC {tex}\Rightarrow{/tex} AF = 2DC
Hence proved.
Since DL = FL (proved above)
{tex}\Rightarrow{/tex} DF = 2DL

0Thank You

Ranveer Singh 6 years, 2 months ago

Bakwas

2Thank You

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