Question

in the below figure ABCD is a quadrilateral and BE is parallel to AC and also meets DC produced at E. show that the area of triangle ADE is equal to the area of the quadrilateral ABCD.

Answer 1

ar(ABCD) = ar(ADC) + ar(ABC).............(i)

ar(ADE) = ar(ADC) + ar(ACE)................(ii)

so, ar(ABCD) = ar(ADE) if ar(ABC) = ar(ACE)

In triangle ABC and ACE,

They have a common base i.e. AC and they have a same height because they are between the same parallel.

Therefore, ar(ABC)=ar(ACE)............(iii)

So, from (iii) we can say that ar(ABCD)=ar(ADE)

Hope you understood. By the way this is my first answer in BRAINLY.

 

Answer 2

Ar. triangle ABC=Ar.triangle ACE(lie on same base AC and between same parallel AC and BE)
so,Ar.ABC+Ar.ADC=Ar.ACE+Ar.ADC
so,Ar.ADE=Ar.ABCD