Question

QUESTION:-IN THE GIVEN FIGURE ABCD IS A QUADRILATERAL AND BE||AC AND ALSO BE MEETS DC PRODUCED AT E . SHOW THAT AREA OF ΔADE IS EQUAL TO THE AREA OF THE 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.

Answer 2

Answer:

Given that BE II AC

⇒ ABEC is a parallelogram, of which opposite sides are parallel.

Δ ABC and Δ ACE lie on the same base AC and between the same parallel AC and BE.

∴ Area(Δ ABC) = Area(Δ ACE)

By adding area(Δ ADC) to both sides

⇒ Area(Δ ABC) + Area(Δ ADC) = Area(Δ ACE) + Area(Δ ADC)

⇒ Area (Parallelogram ABCD) = Area(Δ ADE)