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Internal bisector of $\angle A$ of $\triangle A B C$ meets side $B C$ 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 $a, b$ and $c$ represent sides of $\triangle A B C$, then
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$A E$ is $\mathrm{HM}$ of $b$ and $c$
,
$A D=\frac{2 b c}{b+c} \cos \frac{A}{2}$
,
$E F=\frac{4 b c}{b+c} \sin \frac{A}{2}$
,
the $\triangle A E F$ is isosceles
$A E$ is $\mathrm{HM}$ of $b$ and $c$
,
$A D=\frac{2 b c}{b+c} \cos \frac{A}{2}$
,
$E F=\frac{4 b c}{b+c} \sin \frac{A}{2}$
,
the $\triangle A E F$ is isosceles
We have, $\triangle A B C=\triangle A B D+\triangle A C D$
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} b c \sin A=\frac{1}{2} c A D \sin \frac{A}{2}+\frac{1}{2} b A D \sin \frac{A}{2} \\
& \Rightarrow \quad A D=\frac{2 b c}{b+c} \cos \frac{A}{2}
\end{aligned}
$$
Again, $\quad A E=A D \sec \frac{A}{2}=\frac{2 b c}{b+c}$
$\therefore A E$ is $\mathrm{HM}$ of $b$ and $c$.
$$
\begin{aligned}
E F & =E D+D F=2 D E=2 A D \tan \frac{A}{2} \\
& =2 \frac{2 b c}{b+c} \cos \frac{A}{2} \tan \frac{A}{2}=\frac{4 b c}{b+c} \sin \frac{A}{2}
\end{aligned}
$$
As $A D \perp E F$ and $D E=D F$ and $A D$ is bisector.
$\Rightarrow \triangle A E F$ is isosceles.
$$
\begin{aligned}
& \Rightarrow \frac{1}{2} b c \sin A=\frac{1}{2} c A D \sin \frac{A}{2}+\frac{1}{2} b A D \sin \frac{A}{2} \\
& \Rightarrow \quad A D=\frac{2 b c}{b+c} \cos \frac{A}{2}
\end{aligned}
$$
Again, $\quad A E=A D \sec \frac{A}{2}=\frac{2 b c}{b+c}$
$\therefore A E$ is $\mathrm{HM}$ of $b$ and $c$.
$$
\begin{aligned}
E F & =E D+D F=2 D E=2 A D \tan \frac{A}{2} \\
& =2 \frac{2 b c}{b+c} \cos \frac{A}{2} \tan \frac{A}{2}=\frac{4 b c}{b+c} \sin \frac{A}{2}
\end{aligned}
$$
As $A D \perp E F$ and $D E=D F$ and $A D$ is bisector.
$\Rightarrow \triangle A E F$ is isosceles.
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