Chapter 5.7, Problem 9PSA

To determine

To prove:That the quadrilateral $DMEQ$ is a parallelogram.

Explanation of Solution

Given Information:

  $DM\parallel QR$

  $ME\parallel PQ$

  $PM=MR$

Formula used:

Basic proportionality theorem

Proof:

  

In $\Delta PQR,DM\parallel QR$

Using Basic proportionality theorem, $DM$ divides the sides $PQ\text{and}PR$ in the same ratio

  $\Rightarrow \frac{PD}{DQ}=\frac{PM}{MR}$

As $PM=MR,$ we have

  $\begin{array}{l}\Rightarrow \frac{PD}{DQ}=1\\ \Rightarrow PD=DQ\mathrm{...}\left(\text{i}\right)\end{array}$

Therefore, $D$ is the midpoint of $PQ$

  $\Rightarrow DM=\frac{1}{2}QR\mathrm{...}\left(\text{ii}\right)$

In $\Delta PQR,ME\parallel PQ$

Using Basic proportionality theorem, $ME$ divides the sides $PR\text{and}QR$ in the same ratio

  $\Rightarrow \frac{RM}{MP}=\frac{RE}{EQ}$

As $PM=MR,$ we have

  $\begin{array}{l}\Rightarrow \frac{RE}{EQ}=1\\ \Rightarrow RE=EQ\end{array}$

Therefore, $E$ is the midpoint of $QR$

  $\Rightarrow ME=\frac{1}{2}PQ\mathrm{...}\left(\text{iii}\right)$

  $\begin{array}{l}\text{Now},QR=RE+EQ\\ \Rightarrow QR=2ER\\ \Rightarrow RE=EQ=\frac{1}{2}QR\mathrm{...}\left(\text{iv}\right)\end{array}$

From (ii), we get

  $DM=QE\mathrm{...}\left(\text{v}\right)$

Now, $PQ=PD+DQ$

  $\Rightarrow PD=DQ=\frac{1}{2}PQ$

  $\Rightarrow DQ=ME\mathrm{...}\text{From}\left(\text{iii}\right)$

Thus, in quadrilateral $DMEQ,DM\parallel QE\text{and}DQ\parallel ME$

Also, $DM=QE\text{and}DQ=ME$

Hence, $DMEQ$ is a parallelogram.