Question

$E$ is the midpoint of diagonal $BD$ of a parallelogram $ABCD$. If the point $E$ is joined to a point $F$ on $DA$ such that $DF=\frac{1}{3}DA$, if the ratio of the area of $\Delta DEF$ to the area of quadrilateral $ABEF$ is $1:K$ then find $K$ ?

Answer 1

$ABCD$ is a parallelogram.E is the midpoint of $BD$.$E$ is joined to $F$,where $FD=\frac{1}{3}DA$

Draw $BP\perp AD$ and $EQ\perp AD$

$BP\parallel EQ$

In $BPD$,

$E$ is the midpoint of $BP$and$EQ\parallel BP$

$Q$ is the midpoint of $BD$

$EQ=\frac{1}{2}BP$

$AreaofABD=\frac{1}{2}\ast AD\ast BP$

$AreaofEFD=\frac{1}{2}\ast DF\ast EQ$

$\frac{AreaofEFD}{AreaofABD}=\frac{\frac{1}{2}\ast DF\ast EQ}{\frac{1}{2}\ast AD\ast BP}=\frac{\frac{1}{2}\ast \frac{1}{3}AD\ast \frac{1}{2}BP}{\frac{1}{2}\ast AD\ast BP}=\frac{1}{6}$

$AreaofEFD=\frac{1}{6}AreaofABD$

$AreaofquadrilateralABEF=AreaofABD-AreaofEFD=AreaofABD-\frac{1}{6}AreaofABD=\frac{5}{6}AreaofABD$

$\frac{AreaofEFD}{AreaofquadrilateralABEF}=\frac{\frac{1}{6}AreaofABD}{\frac{5}{6}AreaofABD}=\frac{1}{5}$

The ratio is $1:5$