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.
Answers
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:
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)