Question

In the adjoining figure, CE || AD and CF || BA. Prove that ar(∆CBG) = ar(∆AFG).

Answer 1

$\mathrm{ar}\left(△\mathrm{CFA}\right)=\mathrm{ar}\left(△\mathrm{CFB}\right)$ (Triangles on the same base CF and between the same parallels CF || BA will be equal in area)
$$\begin{gathered} \Rightarrow \mathrm{ar}\left(△\mathrm{CFA}\right)-\mathrm{ar}\left(△\mathrm{CFG}\right)=\mathrm{ar}\left(△\mathrm{CFB}\right)-\mathrm{ar}\left(△\mathrm{CFG}\right) \\ \Rightarrow \mathrm{ar}\left(△\mathrm{AFG}\right)=\mathrm{ar}\left(△\mathrm{CBG}\right) \end{gathered}$$
Hence Proved