Given: A triangle ABC in fig. (i) in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.
To Prove: AD/DB=AE/EC
Construction: Join BE and CD and then draw DM ⊥ AC and EN ⊥ AB.

Now, ΔBDE and ΔDEC are on the same base DE and between the same parallel lines BC and DE.
So, ar(ΔBDE) = ar(ΔDEC) ..(iii)
