Question

Internal bisector of $\angle$ A of AABC m eets side BC at D. A line drawn through D perpendicular to AD in tersects the side AC a t E an d side AB at F. If a , b, c represent sides of $\triangle$ ABC, then

• A. AE is HM of b and c
• B. AD = $\frac{ 2bc}{ b + c} cos \frac{A}{2}$
• C. EF = $\frac{ 4bc}{ b + c} sin \frac{A}{2}$
• D. $\triangle$ AEF is isosceles.

Answer 1

Answer: (D)

$\triangle$ AEF is isosceles.

Answer 2

Since, $\triangle ABC = \triangle ABD + \triangle ACD$ $\Rightarrow \frac{1}{2} bc \, sin \, A = \frac{1}{2} c \, AD \, sin \frac{A}{2} + \frac{1}{2} b \, AD \, sin \, \frac{A}{2}$ $\Rightarrow AD = \frac{ 2bc }{ b + c} \, cos \, \frac{A}{ 2}$ Again, AE = AD sec $\frac{A}{2} = \frac{ 2 bc }{ b + c}$ $\Rightarrow$ AE is HM of b and c EF = ED +D F = 2DE =2 AD tan $\frac{A}{2}$ = 2 $\frac{ 2bc }{ b + c} cos \frac{A}{2} tan \frac{A}{2} = \frac{ 4 bc }{ b + c} sin \frac{A}{2}$ Since, AD $\perp$ EF and DE = DF and AD is bisector. $\Rightarrow \triangle$ AEF is isosceles. Hence, (a), (b), (c), (d) are correct answers.