Question

ABCD is a quadrilateral and BE||AC and also BE meets DC produced at E. show that area of triangle ADE is equal to the area of quadrilateral ABCD.​

Answer 1

Answer:

Consider △AOC and △AOC

∠EBC=∠BCA and ∠BEC=∠ECA as they are alternate angles

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

∠AOC=∠BOE

So, △AOC and △AOC are similar by AAA

And also,

∠AOB=180

∘

−∠AOC=∠COE

Now, consider △AOB and △EOC

sides AO is proportional to OE

sides CO is proportional to OB

and ∠AOB=∠COE

Hence, △AOB and △EOC are similar by SAS

So,

∠ABC=∠BCE and ∠BAE=∠AEC property of similar triangles

So, AB∣∣CE

Hence, ABEC is a parallelogram.