Question

ABCD is a parallelogram and E is the midpoint of BC. The side DC is extended such that it meets at AE when extended at F. Prove that DF = 2DC.​

Answer 1

$\huge \mathcal {\fcolorbox{red}{gray}{\green{A}\pink{N}\orange{S}\purple{W}\red{E}\blue{R}{!}}}$

Solution :-

in the figure

△DCE and BFE

any DEC = any BEF (vertically opp any)

EC=BE (E is the mid point)

∠DCB=∠EBF (alternate angle DC parallel to AF)

So △DCE congruent to △BFE

Therefore DC=BF ...(1)

now CD = AB (ABCD is a parallelogram)

soAF=AB+BF

=AB+DC from (1)

=AB+AB

=2AB

HOPE SO IT WILL HELP.....