Question

In a parallelogram $\mathrm{ABCD}$, $E$ 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

Step $1$: Find the relation between $DP$ and $DQ$.

Given: In $▱\mathrm{ABCD}$, $E$ is the mid-point of $AB$. $F$ is the mid-point of $CD$.

$$\begin{gathered} \begin{array}{cc}& AB\parallel CD\\ \Rightarrow & \frac{AB}{2}\parallel \frac{CD}{2}\\ \Rightarrow & AE\parallel FC\end{array} \\ \begin{array}{cc}& AB=CD\\ \Rightarrow & \frac{AB}{2}=\frac{CD}{2}\\ \Rightarrow & AE=FC\end{array} \end{gathered}$$

Since two sides are equal and parallel $AEFC$ is a parallelogram.

In $△DQC$ and $△DPF$,

$\angle D=\angle D$ $[\because$Common angle$]$

$\angle DPF=\angle DQC$ $[\because$Alternate angles$]$

$\angle DFP=\angle DCQ$ $[\because$Alternate angles$]$

By AAA similarity, $△DQC~△DPF$

$\Rightarrow \frac{DP}{DQ}=\frac{DF}{CD}$

$\Rightarrow \frac{DP}{DQ}=\frac{1}{2}$ $[\because F$ is the midpoint of $CD]$

$\Rightarrow P$ is the midpoint of $DQ$.

From the figure,

$\Rightarrow DP=PQ$………………..$\left(1\right)$

Step $2$: Find the relation between $BP$ and $BQ$.

In $△ABP$ and $△QBE$,

$\angle B=\angle B$ $[\because$Common angle$]$

$\angle BQE=\angle BPA$ $[\because$Alternate angles$]$

$\angle QEB=\angle PAB$ $[\because$Alternate angles$]$

By AAA similarity, $△ABP~△QBE$

$\Rightarrow \frac{BQ}{BP}=\frac{BE}{AB}$

$\Rightarrow \frac{BQ}{BP}=\frac{1}{2}$ $[\because E$ is the midpoint of $CD]$

$\Rightarrow Q$ is the midpoint of $BP$.

From the figure,

$\Rightarrow PQ=BQ$………………..$\left(2\right)$

Step $3$: Prove the given condition.

From equation $\left(1\right)$ and $\left(2\right)$,

$DP=PQ=BQ$

Therefore, from the figure, we can say that, $AF$ and $EC$ trisect the diagonal $BD$.

Hence proved that the line segments $AF$ and $EC$ trisect the diagonal $BD$.