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=1/2(AB+CD)

Name: 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=1/2(AB+CD)
Uploaded: Fri, 2021-01-08 01:20
Description: 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=1/2(AB+CD)

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 $EF=\frac{1}{2}(AB+CD)$

Answers (1)

Given: ABCD is a trapezium in which $AB\parallel CD$ , E and F are the mid-points or sides of 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 the mid-point theorem, O is the mid-point of BD.
$EO=\frac{1}{2}AB$ ……(1)
In $\triangle BDC$ , $OF \parallel CD$ and O is the mid-point of BD
$OF=\frac{1}{2}CD$ …..(2)
Adding 1 and 2, we get
$EO+OF=\frac{1}{2}AB+\frac{1}{2}CD$
$EF=\frac{1}{2}(AB+CD)$