Properties of Triangles 1 Question 15
15. 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 $a, b, c$ represent sides of $\triangle A B C$, then
$(2006,5$ M)
(a) $A E$ is HM of $b$ and $c$
(b) $A D=\frac{2 b c}{b+c} \cos \frac{A}{2}$
(c) $E F=\frac{4 b c}{b+c} \sin \frac{A}{2}$
(d) $\triangle A E F$ is isosceles
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Answer:
Correct Answer: 15. $n=$
Solution:
- Since, $\triangle A B C=\triangle A B D+\triangle A C D$
$\Rightarrow \quad \frac{1}{2} b c \sin A=\frac{1}{2} c A D \sin \frac{A}{2}+\frac{1}{2} b A D \sin \frac{A}{2}$
Again, $A E=A D \sec \frac{A}{2}=\frac{2 b c}{b+c}$
$\Rightarrow \quad A E$ is HM of $b$ and $c$.
$$ \begin{aligned} E F & =E D+D F=2 D E=2 A D \tan \frac{A}{2} \\ & =2 \frac{2 b c}{b+c} \cos \frac{A}{2} \tan \frac{A}{2}=\frac{4 b c}{b+c} \sin \frac{A}{2} \end{aligned} $$
Since, $A D \perp E F$ and $D E=D F$ and $A D$ is bisector.
$\Rightarrow \triangle A E F$ is isosceles.
Hence, (a), (b), (c), (d) are correct answers.
