Chapter 5.5, Problem 33WE

To determine

To find: the best name for quadrilateral $JKLM$ .

Answer to Problem 33WE

Parallelogram

Explanation of Solution

Given:

  $\overrightarrow{VE}$ and $\overrightarrow{FG}$ are congruent.

  $J,K,L,$ and $M$ are midpoint of $\overrightarrow{EF},\overrightarrow{VF},\overrightarrow{VG},$ and $\overrightarrow{EG}$ .

Concept used: Midpoint Theorem that states that a line segment joining the midpoints of two sides of a triangle is parallel to the third side and is equal to half of the third side.

In $\Delta EFG$ , as $J$ and $M$ are midpoints of $\overrightarrow{EF}$ and $\overrightarrow{EG}$ .

So, by Midpoint Theorem,

  $\overrightarrow{JM}=\frac{1}{2}\overrightarrow{FG}$ and $\overrightarrow{JM}$ is parallel to $\overrightarrow{FG}$ ....(i).

In $\Delta VEF$ , as $J$ and $K$ are midpoints of $\overrightarrow{EF}$ and $\overrightarrow{VF}$ .

So, by Midpoint Theorem,

  $\overrightarrow{KJ}=\frac{1}{2}\overrightarrow{VE}$ and $\overrightarrow{KJ}$ is parallel to $\overrightarrow{VE}$ ....(ii).

In $\Delta VEG$ , as $L$ and $M$ are midpoints of $\overrightarrow{VG}$ and $\overrightarrow{EG}$ .

So, by Midpoint Theorem,

  $\overrightarrow{LM}=\frac{1}{2}\overrightarrow{VE}$ and $\overrightarrow{LM}$ is parallel to $\overrightarrow{VE}$ ....(iii).

In $\Delta VFG$ , as $K$ and $L$ are midpoints of $\overrightarrow{VF}$ and $\overrightarrow{VG}$ .

So, by Midpoint Theorem,

  $\overrightarrow{KL}=\frac{1}{2}\overrightarrow{FG}$ and $\overrightarrow{KL}$ is parallel to $\overrightarrow{FG}$ ....(iv).

Now from (i) and (iv)

  $\overrightarrow{JM}=\frac{1}{2}\overrightarrow{FG}=\overrightarrow{KL}$ and $\overrightarrow{JM}\parallel \overrightarrow{FG}\parallel \overrightarrow{KL}$

  $\Rightarrow \overrightarrow{JM}=\overrightarrow{KL}$ and $\overrightarrow{JM}$ is parallel to $\overrightarrow{KL}$ ....(v).

Now from (ii) and (iii)

  $\overrightarrow{LM}=\frac{1}{2}\overrightarrow{VE}=\overrightarrow{KJ}$ and $\overrightarrow{LM}\parallel \overrightarrow{VE}\parallel \overrightarrow{KJ}$

  $\Rightarrow \overrightarrow{LM}=\overrightarrow{KJ}$ and $\overrightarrow{LM}$ is parallel to $\overrightarrow{KJ}$ ....(vi).

Hence from (v) and (vi), $JKLM$ is a parallelogram.

Conclusion:

Therefore, $JKLM$ is a parallelogram.