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

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Solution.
Given: ABCD is a trapezium in which AB\parallel CD, E and F are the mid-points or sides AD and BC.


Constructions: Joint BE and produce it to meet CD at G.
Draw BOD which intersects EF at O
To Prove: EF \parallel AB  and EF=\frac{1}{2}(AB+CD)

Proof: In \triangle GCB, E and F are respectively the mid-points of BG and BC, then by mid-point theorem.
EF \parallel GC
But, GC\parallel AB or  CD \parallel AB           {given}
\therefore EF\parallel AB
In
\triangle ADB, AB \parallel EO and E is the mid-point of AD. Then by mid-point theorem, O is mid-point of BD.
EO=\frac{1}{2}AB          ……(1)
In
\triangle BDC, OF \parallel CDand O is the mid-point of BD
OF=\frac{1}{2}CD         …..(2)
Adding 1 and 2, we get
EO+OF=\frac{1}{2}AB+\frac{1}{2}CD
EF=\frac{1}{2}(AB+CD)

Hence Proved

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