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

Hint: Here, we will use the relation between areas of triangles which have the same base and lie between the same parallel lines. That is, the area of two triangles with the same base and lying between the same parallel lines is always equal.

Complete Step-by-step Solution
Given: ABCDE is a pentagon and $AC\parallel BF$.
We know that from the diagram in the triangle ACB and ACF, the sides AC and BF are parallel to each other and both the triangles lie on the same base AC. So, by the theorem of triangle we can say that the area of ACB and ACF are equal that is,
$ar(ACB) = ar(ACF)$
Therefore, the required relation is proved that is $ar(ACB) = ar(ACF)$.
(ii)We can say that $ar(ACB) = ar(ACF)$ by using relation from part (i).
Now, we will add $ar(ACDE)$ on both sides of the relation $ar(ACB) = ar(ACF)$, we get,
$\begin{array}{c}
ar(ACB) + ar(ACDE) = ar(ACF) + ar(ACDE)\\
ar(ABCDE) = ar(AEDF)
\end{array}$
Therefore, the required relation is proved that is $ar(ABCDE) = ar(AEDF)$.

Note: No other two line segments are parallel here, so it is the only approach you should follow. Otherwise, you would not be able to apply the property that the area of two triangles with the same base and lying between the same parallel lines is always equal.