Question

The line segment joining the mid - points of any two sides of a triangle is parallel to the third side and equal to half of it.
In the given figure, the side AC of $\Delta ABC$ is produced of E such that $CE=\frac{1}{2}AC$. If D is the mid - points of BC and ED produced meets AB at F and CP, DQ are drawn parallel to BA, then FD =

• A. $\frac{1}{2}FE$
• B. $\frac{1}{3}FE$
• C. $FE$
• D. $\frac{1}{4}FE$

Answer 1

The correct option is B $\frac{1}{3}FE$

In $\Delta ABC$, D is the mid - point of BC and $DQ\parallel BA$.
$\therefore$ Q is the mid - point of AC.
[Converse of mid-point theorem]
$\Rightarrow AQ=QC$
Now, $FA\parallel DQ\parallel PC$, and AQC is the transversal such that AQ =QC and FDP is the other transversal on them.
$\therefore$ FD =DP ….(1) [By intercept theorem]
Now, $EC=\frac{1}{2}AC=QC$
$\therefore$ In EQD, C is the mid-point of EQ and $CP\parallel DQ$
$\therefore$ P must be the mid-point of DE.
$\therefore$ DP = PE
Thus, FD = DP = PE [From (1) and (2)]
Hence, $FD=\frac{1}{3}FE$