Question

The side $AC$ of a triangle $ABC$ is produced to point $E$ so that $CE=\frac{1}{2}$ $AC$. $D$ is the mid-point point of $BC$ and $ED$ produced meets $AB$ at F and CP and DQ are parallel to AB .then $FD=\frac{1}{5}FE$ ?

• A. True
• B. False

Answer 1

The correct option is B False

$Here\text{D is the midpoint of BC and DP is parallel to AB}$
$\text{therefore P is the midpoint of AC and PD=}\frac{AB}{2}$
$From\text{triangle}\text{AEF}$
$AE||PD||CRandAP=\frac{AE}{3}$
$HenceDF=\frac{EF}{3}$
$Hencethegivenstatementisfalse$