Question

ABCD is a trapezium in which E & F are mid-points of the sides AD & BC. Prove that EF is ll to AB & EF= 1/2 (AB+CD)

Answer 1

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$SOLUTION$ => 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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Sumit Tiwari ##
Azamgarh #### (UP)