Question

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

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

Answer 1

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

The condition is depicted in the following figure:

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