$a r (△ A C B) = a r (△ A C F)$

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Solution

In figure ABCDE is a pentagon . A line through B parallel to AC meet DC produced at F.
Then AC parallel to BF
Given ABCDE is a pentagon then AB parallel to DC and DC extended up to F
Then AB parallel to CF
The $Δ A C B$ and $A C F$ are same base AC and between two parallel line AC and BF
$∴ a r e a (Δ A C B) = a r e a (Δ A C F)$

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\begin{matrix} In Fig. ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. Show that \end{matrix}

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ar(△ACB)=ar(△ACF)

$a r (△ A C B) = a r (△ A C F)$