Question

State true or false:

In triangle $ABC$. $D$ and $E$ are points on side $AB$ such that $AD=DE=EB$. Through $D$ and $E$ lines are drawn parallel to $BC$ which meet side $AC$ at points $F$ and $G$ respectively. Through $F$ and $G$ lines are drawn parallel to $AB$ which meet side $BC$ at points $M$ and $N$ respectively, then

$BM=MN=NC$.

• A. True
• B. False

Answer 1

The correct option is B False

Given: $△ABC$, D and E are points on AB, such that, $AD=DE=EB$ and F and G are points on AC such that, $DF\parallel EG\parallel BC$
Also, M and N are points on BC such that $FM\parallel GM\parallel AB$
Now, $DF\parallel EG\parallel BC$
hence, by basic proportionality theorem
$AD:DE:EB=AF:FG:GC$
Hence, $AF=FG=GC$
Similarly, $GN\parallel FM\parallel AB$
Hence, by basic proportionality theorem
$AF:FG:GC=BM:MN:NC$
hence, $BM=MN=NC$