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 the side AB at F. If $a,b,c$ represent sides of $\Delta ABC$ 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 triangle AEF is isosceles

Answer 1

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

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

We have $\Delta ABC=\Delta ABD+\Delta ACD$

$\Rightarrow \frac{1}{2}$ $bc$ $\sin A=\frac{1}{2}cAD\sin \frac{A}{2}+\frac{1}{2}b\times AD\sin \frac{A}{2}\Rightarrow$ $AD$ $=\frac{2bc}{b+c}\cos \frac{A}{2}$

Again $AE$ $=$ $AD$ $\sec \frac{A}{2}$

$=\frac{2bc}{b+c}\Rightarrow$ $AE$ is HM of $b$ and $c$.

$EF$ $=$$ED+DF$$=2DE=2\times AD$

$\tan \frac{A}{2}=\frac{2\times 2bc}{b+c}\times \cos \frac{A}{2}\times \tan \frac{A}{2}$

$=\frac{4bc}{b+c}\sin \frac{A}{2}$ As $AD$ $\perp$ $EF$ and $DE$$=$$DF$ and $AD$

is bisector $\Rightarrow$ $AEF$ is isosceles.

Hence $A,B,C$ and $D$ are correct answers.