# In figure, ABCD, DCFE and ABFE are parallelograms. Show that ar(ADE) = ar(BCF).

$$\therefore$$Its opposite sides are parallel and equal.
$$\Rightarrow$$ AD = BC …(1)
Now, $$\triangle$$ADE and $$\triangle$$BCF are on equal bases AD = BC [from (1)] and between the same parallels AB and EF.
So, ar($$\triangle$$ADE) = ar($$\triangle$$BCF).