91Ó°ÊÓ

Regular trapezoid theorem states that the diagonals of a regular trapezoid are equal to each other.

In quadrilateral$ABDE$

$AE$is parallel $BD$(given)

Therefore,$ABDE$ is a trapezoid.

$⇒AD=BE$(Diagonals of a regular trapezoid are equal) …. (i).

In quadrilateral$ABCD$

$BC$is parallel to $AD$(given)

$∴ABCD$is a trapezoid.

$⇒AC=BD$(Diagonals of a regular trapezoid are equal) …. (ii).

Since it is given that $\stackrel{¯}{AD}≈\stackrel{¯}{BD}$, therefore, from (i) and (ii) it can be concluded that

$AD=BD$and $AC=BE$.

$∴ACâ‰\dots BE$.

Hence Proved.

Converse of trapezoid theorem states that if diagonals of a trapezoid are equal to each other then it is a regular trapezoid.

In quadrilateral$ADBE$

$AE$is parallel $BD$(given)

Therefore,$ADBE$is a trapezoid.

$⇒AD=BE$(Diagonals of a regular trapezoid are equal) …. (i).

In quadrilateral$ABCD$

$BC$is parallel to $AD$(given)

$∴ABCD$is a trapezoid.

$⇒AC=BD$(Diagonals of a regular trapezoid are equal) …. (ii).

Since it is given that $\stackrel{¯}{AD}≈\stackrel{¯}{BD}$, therefore, from (i) and (ii) it can be concluded that

$AD=BD$and $AC=BE$.

In quadrilateral $ABCE$,$AC=BE$ (proved above)

Therefore, $EC$is parallel to $AB$. (by the converse of regular trapezoid theorem)