Question

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

Answer 1

Given:

In parallelogram $ABCD$,

$E$ is midpoint on $AB$.

$F$ is midpoint on $CD$.

$AF,EC$ intersects $BD$ at $P,Q$ respectively.

To prove:

$AF$ and $EC$ trisect the diagonal $BD$.

Proof:

In parallelogram $ABCD$,

$AB\parallel DC$

$\Rightarrow AE\parallel FC$ [Since, parts of the parallel lines are parallel]---(a)

$E$ is midpoint on $AB$ $\Rightarrow AE=\frac{1}{2}AB$

$F$ is midpoint on $DC$ $\Rightarrow AF=\frac{1}{2}DC$

Since, $ABCD$ is a parallelogram,

$AB=DC$

$\Rightarrow \frac{1}{2}AB=\frac{1}{2}DC$

$\Rightarrow AE=DF$ ---(b)

From (a) and (b)

$AE=DF$ and $AE\parallel FC$

$\Rightarrow AF=EC$. [Since, distances between equal and the parallel sides]

Hence, $AECF$ is parallelogram.----(1)

Converse of Mid-point theorem: In a triangle if a line passing through mid-point of one side and parallel to second side then it will bisects the third side.

Now, In $\Delta BAP$

$E$ is midpoint of $AB$

$EQ\parallel AP$ [Since, $AECF$ is parallelogram]

By Converse of Mid-point theorem,

$Q$ is mid-point of $BP$.

$\Rightarrow BQ=PQ$----(2)

Now, In $\Delta CDQ$

$F$ is midpoint of $CD$.

$FP\parallel CQ$

By Converse of Mid-point theorem,

$P$ is midpoint of $DQ$.

$\Rightarrow PQ=DP$ --(3)

From (2) and (3) $BQ=PQ=DP$

i.e., diagonal $BD$ divides into three equal parts.

So, $AF,EC$ trisects the diagonal $BD$.

Hence, proved.