$The sides BC,CA and AB of Δ A B C have been produced to D ,E and respectively ∠ B A E + ∠ C B F + ∠ A C D = ? .$

$(a) 240^{\circ} (b) 300^{\circ} (c) 320^{\circ} (d) 360^{\circ}$

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Solution

ANSWER:
(d) $360^{\circ}$
We have : $∠ 1 + ∠ B A E = 180^{\circ}$ ...i

$∠ 2 + ∠ C B F = 180^{\circ} . . . i i a n d ∠ 3 + ∠ A C D = 180^{\circ}$ ...iii
Adding (i), (ii) and (iii), we get: $∠ 1 + ∠ 2 + ∠ 3 + ∠ B A E + ∠ C B F + ∠ A C D = 540^{\circ}$
⇒ $180^{\circ} + ∠ B A E + ∠ C B F + ∠ A C D = 540^{\circ} ∵ ∠ 1 + ∠ 2 + ∠ 3 = 180^{\circ} \Rightarrow ∠ B A E + ∠ C B F + ∠ A C D = 360^{\circ}$

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