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Applying Menelaus' theorem to ΔBCF with AD as the transversal, we have

= 1

But BD/DC = 1/2 (as BD = DE = EC) and CA/AF = 2/1 (as CF = FA).

Hence we have BP = PF.

Again applying Menelaus' Theorem to ΔBCF with AE as the transversal we have = 1

But BE/EC = 2/1 and CA/AF = 2/1

Hence 4FQ = QB.

Suppose FQ= x unit. The QB = 4x unit. That is BF = 5x unit. Since BP = PF hence each is 2.5x unit.

Thus PQ = 2.5x - x = 1.5x unit

Hence =

Also

Thus = = \( \frac {1.5x}{5x} \frac {1}{2}\) = = ...(1)

Again \( \frac {\triangle ADE}{\triangle ABC}\) = (as DE/BC = 1/3)

Thus \( \frac {\triangle ADE}{\triangle ABC} - \frac {\triangle APQ}{\triangle ABC} = \frac {1}{3} - \frac {1.5}{10}\)

...(2)

Using (1) and (2) we have

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