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JEE Advanced · Mathematics · 17. Properties of Triangles

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 \(a, b\) and \(c\) represent sides of \(\triangle A B C\), then

  1. A
    \(A E\) is \(\mathrm{HM}\) of \(b\) and \(c\)
  2. B
    \(A D=\frac{2 b c}{b+c} \cos \frac{A}{2}\)
  3. C
    \(E F=\frac{4 b c}{b+c} \sin \frac{A}{2}\)
  4. D
    the \(\triangle A E F\) is isosceles
Verified Solution

Answer & Solution

Correct Answer

(A)
\(A E\) is \(\mathrm{HM}\) of \(b\) and \(c\)

Step-by-step Solution

Detailed explanation

We have, \(\triangle A B C=\triangle A B D+\triangle A C D\)
\[
\begin{aligned}
& \Rightarrow \frac{1}{2} b c \sin A=\frac{1}{2} c A D \sin \frac{A}{2}+\frac{1}{2} b A D \sin \frac{A}{2} \\
& \Rightarrow \quad A D=\frac{2 b c}{b+c} \cos \frac{A}{2}
\end{aligned}
\]
Again, \(\quad A E=A D \sec \frac{A}{2}=\frac{2 b c}{b+c}\)
\(\therefore A E\) is \(\mathrm{HM}\) of \(b\) and \(c\).
\[
\begin{aligned}
E F & =E D+D F=2 D E=2 A D \tan \frac{A}{2} \\
& =2 \frac{2 b c}{b+c} \cos \frac{A}{2} \tan \frac{A}{2}=\frac{4 b c}{b+c} \sin \frac{A}{2}
\end{aligned}
\]
As \(A D \perp E F\) and \(D E=D F\) and \(A D\) is bisector.
\(\Rightarrow \triangle A E F\) is isosceles.

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