91Ó°ÊÓ

It is known that a parallelogram is a quadrilateral whose both pairs of opposite sides are parallel.

From the above figure, it can be observed that the angles $∠XHG$and$∠HEF$ are the corresponding angles that are formed when the lines $\stackrel{¯}{HG}$and$\stackrel{¯}{EF}$ are cut by the transversal $\stackrel{¯}{XE}$.

It can be observed that,

$∠XHG=∠HEF=70°$

That is,

$∠XHGâ‰\dots ∠HEF$

Thus, by postulate 11 which states that the relation between two lines is parallel only when a separate line travels through these two lines and the angles that are corresponding will be congruent, it can be said that the lines $\stackrel{¯}{HG}$and$\stackrel{¯}{EF}$ are parallel.

The angles $∠GFY$and $∠HEF$are the corresponding angles that are formed when the lines $\stackrel{¯}{HE}$and $\stackrel{¯}{GF}$are cut by the transversal $\stackrel{¯}{EY}$.

It can be observed that,

$∠GFY=∠HEF=70°$

That is, $∠GFYâ‰\dots ∠HEF$

Thus, by postulate 11 which states that the relation between two lines is parallel only when a separate line travels through these two lines and the angles that are corresponding will be congruent, it can be said that the lines $\stackrel{¯}{HE}$and $\stackrel{¯}{GF}$are parallel.

That is, for the quadrilateral EFGH whose both pairs of opposite sides are parallel.

So, it can be said that EFGH is a parallelogram.