91Ó°ÊÓ

Consider the diagram.

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.

In $\mathrm{Δ}EFG$, as J and M are midpoints of $\stackrel{→}{EF}$and $\stackrel{→}{EG}$.

So, by Midpoint Theorem,

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

In $\mathrm{Δ}VEF$, as J and K are midpoints of $\stackrel{→}{EF}$and $\stackrel{→}{VF}$.

So, by Midpoint Theorem,

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

In $\mathrm{Δ}VEG$, as L and M are midpoints of$\stackrel{→}{VG}$ and $\stackrel{→}{EG}$.

So, by Midpoint Theorem,

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

In $\mathrm{Δ}VFG$, as K and L are midpoints of $\stackrel{→}{VF}$and $\stackrel{→}{VG}$.

So, by Midpoint Theorem,

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

Now from (i) and (iv)

$\stackrel{→}{JM}=\frac{1}{2}\stackrel{→}{FG}=\stackrel{→}{KL}$and $\stackrel{→}{JM}âˆ\yen \stackrel{→}{FG}âˆ\yen \stackrel{→}{KL}$

$⇒\stackrel{→}{JM}=\stackrel{→}{KL}$and $\stackrel{→}{JM}$is parallel to $\stackrel{→}{KL}$. …. (v)

Now from (ii) and (iii)

$\stackrel{→}{LM}=\frac{1}{2}\stackrel{→}{VE}=\stackrel{→}{KJ}$and $\stackrel{→}{LM}âˆ\yen \stackrel{→}{VE}âˆ\yen \stackrel{→}{KJ}$

$⇒\stackrel{→}{LM}=\stackrel{→}{KJ}$and $\stackrel{→}{LM}$is parallel to $\stackrel{→}{KJ}$ …. (vi)

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

Therefore, $JKLM$is a parallelogram.