Question

In fig, $ABCD$ is a quadrilateral. $BE\parallel AC.BE$ meets $DC$ (produced) at $E.AE$ and $BC$ intersect at $O$. Which one is the correct answer from the following?

• A. $ABEC$ is a parallelogram
• B. $ar\left(△AOC\right)=ar\left(△BOE\right)$
• C. $ar\left(△OAB\right)=ar\left(△OCE\right)$
• D. $ar\left(△ABE\right)=ar\left(△ACE\right)$

Answer 1

The correct option is A $ABEC$ is a parallelogram

Consider $△AOC$ and $△AOC$

$\angle EBC=\angle BCA$ and $\angle BEC=\angle ECA$ as they are alternate angles

Hence, as two angles are equal the third angle is also equal.

$\angle AOC=\angle BOE$

So, $△AOC$ and $△AOC$ are similar by AAA

And also,

$\angle AOB={180}^{\circ }-\angle AOC=\angle COE$

Now, consider $△AOB$ and $△EOC$

sides $AO$ is proportional to $OE$

sides $CO$ is proportional to $OB$

and $\angle AOB=\angle COE$

Hence, $△AOB$ and $△EOC$ are similar by SAS

So,

$\angle ABC=\angle BCE$ and $\angle BAE=\angle AEC$ property of similar triangles

So, $AB||CE$

Hence, ABEC is a parallelogram.