In Figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that
i) ar (ACB) = ar (ACF)
ii) ar (AEDF) = ar (ABCDE).

i) \(\triangle{ACB}\) and \(\triangle{ACF}\) lie on the same base AC and are between The same parallels AC and BF.

\(\therefore \) Area (\(\triangle{ACB}\)) = Area (\(\triangle{ACF}\))


ii) \(\because \) it can be observed that
Area (\(\triangle{ACB}\)) = Area (\(\triangle{ACF}\))

Now adding Area (ACDE) on both side,

\(\Rightarrow \) Area (\(\triangle{ACB}\)) + Area (ACDE) = Area (\(\triangle{ACF}\)) + Area (ACDE)

\(\Rightarrow \) Area (ABCDE) = Area (AEDF)
Hence, proved.