Question

Directions: The following question has four choices, out of which one or more are correct.The internal bisector of $\mathrm{\angle }A$ of a 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 $a,b$ and $c$ represent the sides of $\mathrm{\Delta }ABC$, 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) $\mathrm{△}AEF$ is isosceles

Answer 1

The Correct Option is A, D

Explanation:

$$\begin{array}{l}\text{ We have }\mathrm{\Delta }ABC=\mathrm{\Delta }ABD+\mathrm{\Delta }ACD\\ ⟹\frac{1}{2}b\mathrm{csin}A=\frac{1}{2}cAD\sin \frac{A}{2}+\frac{1}{2}b\times AD\sin \frac{A}{2}\\ ⟹AD=\frac{2bc}{b+c}\cos \frac{A}{2}\\ \text{ Again }AE=AD\sec \frac{A}{2}\\ =\frac{2bc}{b+c}⟹AE\text{ is }HM\text{ of }b\text{ and }c\text{. }\end{array}$$$$\begin{array}{rl}EF=ED+& DF=2DE=2\times ADtan\frac{A}{2}=\frac{2\times 2bc}{b+c}\times \cos \frac{A}{2}\times \tan \frac{A}{2}\\ & =\frac{4bc}{b+c}\sin \frac{A}{2}\end{array}$$As $AD\perp EF$ and $DE=DF$ and $AD$ is bisector $⟹AEF$ is isosceles.