Question

In the figure given below, $ABCD$ is a trapezium in which $AB\parallel DC$. $E$ and $F$ are the midpoint of $AD$ and $BC$ respectively. $DF$ and $AB$ are produced to meet at $G$. Also $AC$ and $EF$ intersect at point $O$. Show that :
$\left(i\right)EO\parallel AB$
$\left(i\right)AO=CO$

Answer 1

$ABCD$ is the given trapezium.

It is given that $E$ and $F$ are the mid-point of the sides $AD$ and $BC$ respectively.

$\left(i\right)$ Consider $△ADG,$

By the converse of midpoint theorem,

$EF\parallel AG$ and $EF=\frac{1}{2}AG$

$\Rightarrow$ $EO\parallel AG$

$\Rightarrow$ $EO\parallel AB$

$\left(ii\right)$ Consider $△ADC$

$EO\parallel AB$ and $AB\parallel DC$

$\Rightarrow$ $EO\parallel DC$

And we know that $E$ is the midpoint of $AD$.

Thus, by basic proportionality theorem, we have, $O$ is the mid-point of $AC$

$\therefore$ $AO=CO$