Question

In the figure (2) given below, DE is drawn parallel to the diagonal AC of de
quadrilateral ABCD to meet BC produced at the point E. Prove that area of quad
A(BCD) = area of A(ABE)​

Answer 1

Since DE is parallel to AC, then

ar(ADC)=ar(ACE) areas of figures between two parallel lines and on same base are equal).

Now, adding ar(ABC) on both sides we get:

ar(ADC)+ar(ABC) =ar(ACE) + ar(ABC)

= ar(ABCD)= ar(ABE)

hence, proved.