Question

40. E and Fare respectively the mid-points of the non-parallel sides AD and BC of a trapezium
ABCD. Prove that EF || AB and EF =half (AB+CD).
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Answer 1

Hey!✔

 ABCD is a trapezium in which AB || DC and E, F are mid points of AD, BC respectively. Join CE and produce it to meet BA produced at G. 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) ∴ ΔEDC ≅ ΔEAG CD  = GA and EC = EG  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)

.....Hope. this helps....

Answer 2

The complete Solution Is In The Attachment

Hope you understood correctly.