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In △ABC, interval angle bisector of ∠A meets side BC in D. DE ⊥ AD meets AC in E and AB in F. Then

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In △ABC, interval angle bisector of ∠A meets side BC in D. DE ⊥ AD meets AC in E and AB in F. Then

(a) AE is HM of b and c

(b) AD = \(\frac{2bc}{b +c}\) cos\(\frac A2\)

(c) EF = \(\frac{4bc}{b + c}\) sin\(\frac A2\)

(d) △AEF is isosceles

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Hence option a, b, c, d are correct.

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