Question

in figure ABCDE is a pentagon a line through B parrallel to AC meets DC produced at F so that ar(triangle ACB )is equal to ar (yriangle ACF ). (b)ar (AEDF) is equal to ar ( ABCDE)

Answer 1

i) To prove ar(ACB) = ar(ACF)
Since, the triangles between the same parallels and on the same base are equal in area.
i.e., ΔACB and ΔACF are on the same base AC and between same parallels AC and BF
∴ ar(ACB) = ar(ACF)

ii)To prove ar(AEDF) = ar(ABCDE)
since, ar(ACB) = ar(ACF) [ Proved]
Adding ar(AEDC) to both sides we get
⇒[ar(ACB) + ar(AEDC)] = [ar(ACF) + ar(AEDC)]
⇒ ar(AEDF) = ar(ABCDE)

Answer 2

Hi friend..

See the attached file

I hope it will help you

✌️☺️