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 $△$ 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. $△$ AEF is isosceles.

Answer 1

Answer: (A), (B), (C), (D)

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