Question

e and f are the mid points of the non parallel side of ad and bc of trapezium abcd prove that ef || abef

Answer 1

Hey!

Here is your answer.

As in the question ABCD is a trapezium in which AB || DC and E, F are mid points of AD, BC respectively.

So join CE and produce it to meet BA produced at G.

Now in ΔEDC and ΔEAG,

ED = EA    ( E is mid point of AD)

∠CED = ∠GEC ( Vertically opposite angles)

∠ECD = ∠EGA ( alternate angles) ( DC||AB, DC||GB and CG transversal)

So obviosly  ΔEDC ≅ ΔEAG

CD  = GA and EC = EG 

Next

In ΔCGB,

E is mid point of CG ( EC = EG proved)

F is a mid point of BC  (given)

∴ By mid point theorem EF ||AB and EF = (1/2)GB.

But GB = GA + AB = CD + AB

Hence EF||AB and EF = (1/2)( AB + CD).
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Hope it helped u understand :).