Question

Internal bisector of $\angle A$ of triangle $ABC$ meets side $BC$ at $D$. A line drawn through $D$ perpendicular to $A D$ intersects the side $A C$ at $E$ and side $A B$ at $F$. If $a, b, c$ represent sides of $\triangle ABC$, then

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

Answer 1

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

The condition is depicted in the following figure:

Now, Area of $\triangle A B C=$ Area of $\triangle A B D+$ Area of $\triangle A C D$
$\left(\sin a \times \frac{1}{2} \times b c\right)=\left(\frac{1}{2} \times c \times A D \times \sin \frac{A}{2}\right)+\left(\frac{1}{2} \times b \times A D \times \sin \frac{A}{2}\right)$
$\sin A \left(\frac{b c}{b+c}\right)= AD \sin \frac{A}{2}$
$AD =\frac{2 b c \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)}{(b+c) \sin \left(\frac{A}{2}\right)}$
$=\left(\frac{2 b c}{b+c}\right) \times\left(\cos \frac{A}{2}\right)$
From $\triangle ADE$, we have
$\cos \left(\frac{A}{2}\right)=\frac{A D}{A E} $
$\Rightarrow AE =\frac{2 b c}{b+c}$
where $AE$ is $HM$ of $b$ and $c$.
From $\triangle ADF$, we have
$\cos \left(\frac{A}{2}\right)=\frac{A D}{A F} $
$\Rightarrow AR = AF \Rightarrow \triangle AEF$
which is an isosceles triangle.
Therefore, ain $\left(\frac{A}{2}\right)=\frac{D F}{A F}$
$\Rightarrow EF =2 DF =2 AF \sin \frac{A}{2}$
$=\frac{4 b c}{b+c} \sin \left(\frac{A}{2}\right)$