Question

$\overline{VE},\overline{FG}$are congruent. $J,K,L,M$are the midpoint of $\overline{EF},\overline{VF},\overline{EG},\overline{VG}$. What name best describes JKLM? Explain.

Short answer

JKLM is a parallelogram.

Step-by-step solution

Step 1. Check the diagram.

Consider the diagram.

Step 2. Apply the concept of midpoint theorem.

Midpoint Theorem 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.

Step 3. Step description.

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 $\overrightarrow{EG}$.

Step 4. Step description.

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.

Therefore, $JKLM$is a parallelogram.