Question

36. In Figure, ABCD is a parallelogram and E is the mid-point of AD. A line through D. drawn
parallel to EB, meets AB produced at F and BC at L. Prove that
i AF = 2DC
ii. DF = 2DL
A
8​

Answer 1

Answer:

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