Q.
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
IIT JEEIIT JEE 2006
Solution:
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.
