Question

The internal bisector of the angle $A$ of the $△ABC$ meets $BC$ at $D$.

A line drawn through $D$ perpendicular to $AD$ intersects the side AC at $E$ and the side $AB$ produced at $F$.

Then which of the following is true?

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

Answer 1

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

The correct options are
A HM of $b$ and $c$ is equal to $AE$
B the $△AEF$ is isosceles
C $EF=\frac{4bc}{b+c}\sin \frac{A}{2}$
D $AD=\frac{2bc}{b+c}\cos \frac{A}{2}$

$\Delta ABC=\Delta ABD+\Delta ACD$
$\Rightarrow \frac{1}{2}bc\sin A=\frac{1}{2}cAD\sin \frac{A}{2}+\frac{1}{2}bAD\sin \frac{A}{2}$
$\Rightarrow AD=\frac{2bc}{b+c}\cos \frac{A}{2}$
$AE=AD\sec \frac{A}{2}=\frac{2bc}{b+c}$
$\Rightarrow AE$ is H.M between $b$ and $c$
$EF=ED+DF=2\times AD\tan \frac{A}{2}=\frac{4bc}{b+c}\sin \frac{A}{2}$
As $AD$ is perpendicular to $EF$, $DE=DF$, $\Delta AEF$ is isosceles