Question

ABCD is a parallelogram. E and F are the mid-Points of BC and AD respectively. Show that the segments BF and DE trisect the diagonal AC. [4 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark
Proof: 2 Marks

$SinceDF=\frac{1}{2}ADandBE=\frac{1}{2}BC\Rightarrow DF=BE$

Also, since $AD\parallel BC$, we can say that $DF\parallel BE.$

Thus, we can conclude that $DFBE$ is a parallelogram and that it’s opposite sides $FB$ and $DE$ are parallel.

Now consider $\Delta CPB$

$E$ is the midpoint of $BC$ and $QE\parallel PB$.

$\Rightarrow Q$ is the midpoint of $CPorCQ=QP$……..(1)
[Converse of midpoint theorem]

Now consider $\Delta ADQ$

$F$ is the midpoint of $AD$ and $DQ\parallel FP$.

$\Rightarrow P$ is the midpoint of $AQ$ or $AP=QP$……..(2) [Converse of midpoint theorem]

Combining (1) and (2), we get,
$CQ=QP$
$AP=QP$
$\Rightarrow AP=PQ=QC$

$\therefore$ The line segments $BF$ and $DE$ trisect the diagonal $AC$.

$\text{Hence proved.}$