Question

ABCD
is trapezium with AD || BC
E is mid-point of AB
F med - point of CD

show
seg E F || side AD
:​

Answer 1

Answer:

AB || CD  (given)

 

E is mid point of AD (given)

 

EF || DC (given)

 

Let us join BD and let EF meets BD at G.

 

since EF || DC and AB || DC, we have EG || AB.

 

In Δ ABD, EG || AB and E is mid point of AD. Hence EG = (1/2)AB and BG = DG (1)

 

similarly we can say GF || DC.

 

in Δ BCD,  FG || CD and G is mid point of BD. Hence GF = (1/2)DC (2)

 

from (1) and (2), we have   EG+GF = EF = (1/2)(AB+DC)

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