Question

ABCD is a trapezium in which side AB is parallel to DC and E is the mid point of side AD. If F is a point on BC such that the line segment EF is parallel to DC, then prove that EF = 1/2(AB+DC)

Answer 1

$ABCD$ is trapezium in which $AB\parallel DC$.

$EF$ is parallel to side $DC$.

Then we have $AB\parallel DC\parallel EF$.

Hence we have also trapezium $ABFE$ and trapezium $EFCD$.

Let $AP$ be the perpendicular to $DC$ and this intersects $EF$ at $Q$.

$AQ$ will be perpendicular to $EF$.

For $△APD$ and $△AQE$ we have $\frac{AD}{EA}=\frac{AP}{AQ}=2$

This gives $AP=2AQ$

i.e, $AQ=QP$

Consider the area we have area $ABCD=$ area $ABFE+$ area $EFCD$

$\left(\frac{1}{2}\right)AP\times (AB+DC)=\left(\frac{1}{2}\right)AQ\times (AB+EF)+\left(\frac{1}{2}\right)QP\times (EF+DC)$

$\Rightarrow AP(AB+DC)=AP\times \frac{AB}{2}+AP\times \frac{EF}{2}+AP\times \frac{EF}{2}+AP\times \frac{DC}{2}$

$\Rightarrow AP\times \frac{AB}{2}+AP\times \frac{DC}{2}=AP\times \frac{EF}{2}+AP\times \frac{EF}{2}$

$\Rightarrow AP\times \frac{(AB+DC)}{2}=AP\times EF$

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