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Answer :
Since, ABCD is a parallelogram [Given]
\(\therefore \)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).