Question

Internal bisector of $\angle$A of a triangle $ABC$ meets the 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 the sides of $\Delta$ABC, then which of the following is correct?

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

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

using formula for area of a triangle

$\Rightarrow \frac{1}{2}bcsinA=\frac{1}{2}cADsin\left(A/2\right)+\frac{1}{2}bADsin\left(A/2\right)$

using cosine formula

$\Rightarrow AD=\frac{2bc}{b+c}cos\frac{A}{2}$
$AE=ADsec\frac{A}{2}=\frac{2bc}{b+c}$

$\Rightarrow$ AE is H.M. between b and c

$EF=ED+DF=2\times ADtan\frac{A}{2}=\frac{4bc}{b+c}sin\frac{A}{2}$

As AD is perp. to EF, DE $=$ DE, $\Delta$AEF is isosceles.

Answer is option A, B, C, D