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Question 12
E and F are respectively the mid-points of the non-parallel sides AD and BC of a trapezium ABCD, Prove that EF||AB and EF=12(AB+CD).

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Solution

Given ABCD is a trapezium in which AB||CD. Also, E and F are respectively the mid-points of sides AD and BC.

Construction:- Join BE and produce it to meet CD produced at G, also draw BD which intersects EF at O.
To prove EF|| AB and EF=12(AB+CD)

Proof :- In ΔGCB,E and F are respectively the mid-points of BG and BC, then by mid-point theorem,

EF||GC

But GC||AB or CD||AB

EF||AB

In ΔADB,AB||EO and E is the mid-point of AD.

Therefore by converse of mid-point theorem, O is mid-point of BD.
Also, EO=12AB

In ΔBDC,OF||CD and O is the mid-point of BD.
OF=12CD [by converse of mid-point theorem]

On adding Eqs. (i) and (ii), we get

EO+OF=12AB+12CDEF=12(AB+CD)
Hence, proved

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