Question

In $\Delta ABC,$ the side $AC$ is produced to $E$ such that $CE=\frac{1}{2}AC$. If $D$ is the midpoint of $BC$ and $ED$ produced meets $AB$ at $F$, and $CP$, $DQ$ are drawn parallel to $BA$, then $FD=\frac{1}{3}FE$.

• A. True
• B. False

Answer 1

The correct option is A True

Given,

$ABC$ is a triangle.

$D$ is midpoint of $BC$ and $DQ$.

They're drawn parallel to $BA$.

Then,

$Q$ is midpoint of $AC$.

$AQ=DC$

Since,

$FA$ parallel to $DQ||PC$.

$AQC$, is a transverse

So, $AQ=QC$ and $FDP$ also a transverse on them.

$FD=DP$ .......(1) [ intercept theorem]

$EC=1/2AC=QC$

Now,

Triangle $EQD$,

Here $C$ is midpoint of $EQ$ and $CP$ which is parallel to $DQ$.

And, $P$ is midpoint of $DE$.

$DP=PE$..........(2)

Therefore,

From (1) and (2)

$FD=DP=PE$

$FD=1/3FE$

Hence, this is the answer.