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In Figure, ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show thati) 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.