Question

$AD$ is the angular bisector of $\angle A$. $DE,$ perpendicular to $AD$, intersects $AC$ at $E$ and meets $AB$ produced at $F$, then

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

Answer 1

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

The correct options are
A $△AEF$ is isosceles
B $AD=\frac{2bc}{b+c}\cos \frac{A}{2}$
C $AE$ is the $H.M.$ of $b$ and $c$

From $△ABC\frac{AD}{\sin B}=\frac{BD}{\sin \frac{A}{2}}$,

And, $\frac{BD}{DC}=\frac{c}{b}$ $[\because AD$ is bisector of $\angle A]$

$\therefore \frac{BD}{BD+DC}=\frac{c}{b+c}\Rightarrow BD=\frac{ca}{b+c}$ $(\because BD+DC=BC=a)$

$\therefore AD=\frac{ca}{b+c}.\frac{\sin B}{\sin \frac{A}{2}}$$=\frac{ca\sin B.2\cos \frac{A}{2}}{(b+c).2\sin \frac{A}{2}\cos \frac{A}{2}}=2.\frac{c}{b+c}.\frac{a}{\sin A}.\sin B.\cos \frac{A}{2}$

$=\frac{2c}{b+c}.\frac{b}{\sin B}\sin B\cos \frac{A}{2}=\frac{2bc}{b+c}\cos \frac{A}{2}.$$\therefore$ (b) holds.

Again, $△AED=△AFD$

$[\because AD=AD,\angle EAD=\angle FAD=\frac{A}{2}\angle ADE=\angle ADF={90}^0]$

$\therefore AE=AF$$\therefore △ABC,$is isosceles, $\therefore$ (a) holds.

In $△AED,\cos \frac{A}{2}=\frac{AD}{AE}$$\Rightarrow AE=\frac{AD}{\cos \frac{A}{2}}=\frac{2bc}{b+c}$

$\therefore AE$ is $H.M.$ of $b$ and $C.$ $\therefore$ (c) holds

$\therefore AE=AF$