Question

$AD$ is internal angle bisector of $△ABC$ at $\angle A$ and $DE$ perpendicular to $AD$ which intersects $AC$ at $E$ and meets $AB$ in $F$, then:

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

Answer 1

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

The correct options are
A $AF=\frac{4bc}{b+c}\sin \frac{A}{2}$
B $AD=\frac{2bc}{b+c}\cos \frac{A}{2}$
C $AE$ is harmonic mean of $b$ and $c$
D $△AEF$ is an isosceles triangle

$△ABC=△ABD+△ACD$
$\frac{1}{2}bc\sin A=\frac{1}{2}AD.c\sin \frac{A}{2}+\frac{1}{2}AD.b\sin \frac{A}{2}$
or $bc\left(2\cos \frac{A}{2}\right)=AD(b+c)$
$\therefore AD=\frac{2bc}{b+c}\cos \frac{A}{2}...\left(1\right)$
Again $AD=AE\cos \frac{A}{2}$ from $△AED$
$\therefore AE=\frac{AD}{\cos \frac{A}{2}}=\frac{2bc}{b+c}$ by (1)
i.e., $AE$ is H.M of $b$ and $c$
Again $EF=ED+DF=2ED=2AD\tan \frac{A}{2}=2.\frac{2bc}{b+c}\cos \frac{A}{2}\tan \frac{A}{2}=\frac{4bc}{b+c}\sin \frac{A}{2}$
$△AED=△AFD$
$AE=AF$ $\Rightarrow$ $△AEF$ is isosceles.