Question

A point E is taken on the side BC of a parallelogram ABCD.AE and DC are produced to meet at F.Prove that ar(∆ADF)=ar(ABFC)

Answer 1

◆Ekansh Nimbalkar◆
Hello friend here is your required answer

Given.ABCD is a parallelogram.A point E is taken on the side BC.AE and DC are produced to meet at F
Prove that ar(∆ADF)=ar(∆ADF).....(1)
As DC||AB,soCF||AB
Since triangles on the same base and between the same parallels are equal in area,so we have ar(∆ACF)=ar(∆BCF).....(2)
Adding(1) and (2),we get
ar(∆ADC)+ar(∆ACF)=ar(∆ABC)+ar(∆BCF)
=ar(∆ADF)=ar(ABFC). Hence proved.

Answer 2

Answer is given - by Ajay Sahu