Question

In a parallelogram $ABCD,E$ and $F$ are the mid-points of sides $AB$ and $CD$ respectively. Show that the line segments $AF$ and $EC$ trisect the diagonal $BD$.

Answer 1

$ABCD$ is a paralleogram

So $AB||CD$

$\Rightarrow AE||CF$ (parts of ||el lines are ||el)

$AB||CD$

$\frac{1}{2}AB=\frac{1}{2}CD$

$\Rightarrow AE=CF$

In $ACFE$

$AE||CF$ and $AE=CF$

one side of opposite sides are equal and ||el thus $ACFE$ is a parralleogram

$\Rightarrow AF||CE$

$\therefore PF||CQ$ and $AP||EQ$

In $△DQC$

$F$ is mid point of $DC$

$PF||CQ$

In $△ABP$

$E$ is mid point as $AB$

$AP||EQ$

[line drawn through mid point of one side of a triangle parallel to another side

, bisect the third side]

$\Rightarrow PQ=DP$ (i) $\Rightarrow PQ=QB$

thus

$DP=PQ=BQ$

Hence proved