Question

Internal bisector of ∠A of triangle ABC meets side BC at D. A line drawn through D perpendicular to AD intersects the side AC at E and side AB at F. If a,b,c represent sides of ∆ ABC, then

(A) AE is HM of b and c

(B) AD = 2bc/(b + c)cos(A/2)

(C) EF = 4bc/(b + c)sin(A/2)

(D)The ∆ AEF is isosceles

Answer 1

Correct option (A, B, C, D)

Explanation:

Since, ∆ ABC = ∆ ABD + ∆ ACD

⇒ 1/2bcsinA = 1/2c AD sin(A/2) + 1/2b AD sin(A/2)

⇒ AD = 2bc/(b + c)cos(A/2)

Again, AE = AD sec(A/2)

⇒ AE is HM of b and c.

EF = ED + DF = 2DE = 2AD tan(A/2)

= 2(2bc/(b + c))cos(A/2)tan(A/2) = 4bc/(b + c)sin(A/2)

Since, AD ⊥ EF and DE = DF and AD is bisector.

⇒ ∆ AEF is isosceles