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

Solve for the required proof

Given:

1. $ABCD$ is a parallelogram
2. $DF=FC=\frac{DC}{2}$
3. $AE=EB=\frac{AB}{2}$

Construction: Join $AF$ and $EC$ which intersect the diagonal $BD$ at point $P$ and point $Q$ respectively.

To prove: $DP=PQ=QB=\frac{BD}{3}$

Proof :

Step 1: Prove that $AEFC$ is a parallelogram

Parallel sides of a parallelogram are equal

$\Rightarrow$ $AB=CD$ [Since, $ABCD$ is a parallelogram]

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

$\Rightarrow$ $AE=FC$ $...\left(i\right)$

As $AE=FC$ and $AE\parallel FC$ we can say that $AEFC$ is a parallelogram, as opposite sides are equal and parallel.

$\Rightarrow$$AF\parallel CE$ opposite sides a of a parallelogram

$\Rightarrow$$FP\parallel CQ$ $...\left(ii\right)$

$\Rightarrow$$AP\parallel EQ$ $...\left(iii\right)$

Step 2: Show that $DP=PQ$

Consider $△DQC$

By converse of midpoint theorem we can say that line parallel to a side passing through the midpoint of other side bisects the third side of the triangle

As $FP\parallel CQ$, $F$ is midpoint of $CD$

We can conclude that $P$ is the midpoint of side $DQ$

$\Rightarrow$$DP=PQ$ $...\left(iv\right)$

Step 3: Show that $BP=PQ$

Consider $△APB$

By converse of midpoint theorem we can say that line parallel to a side passing through the midpoint of other side bisects the third side of the triangle

As $AP\parallel EQ$,$E$ is midpoint of $AB$

We can conclude that $Q$ is the midpoint of side $PB$

$\Rightarrow$$BP=PQ$ $...\left(v\right)$

Step 4: Show that $DP=PQ=QB=\frac{BD}{3}$

From $\left(iv\right),\left(v\right)$

$DP=PQ=QB$

$DP+PQ+QB=BD$

$\Rightarrow$ $3DP=BD$

$\Rightarrow$ $DP=\frac{BD}{3}$

$\Rightarrow$$DP=PQ=QB=\frac{BD}{3}$

Hence, it is proved that $AF$ and $EC$ trisect the diagonal $BD$.