E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD.Prove that EF is parallel to AB and EF=1/2(AB+CD)
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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).
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