17. Internal bisector of $\angle \mathrm { A }$ 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 $\mathrm { a } , \mathrm { b } , \mathrm { c }$ represent sides of $\triangle \mathrm { ABC }$ then\\
(A) AE is HM of b and c\\
(B) $\mathrm { AD } = \frac { 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \cos \frac { \mathrm { A } } { 2 }$\\
(C) $\mathrm { EF } = \frac { 4 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \sin \frac { \mathrm { A } } { 2 }$\\
(D) the triangle AEF is isosceles

Sol. (A), (B), (C), (D).\\
We have $\triangle \mathrm { ABC } = \triangle \mathrm { ABD } + \triangle \mathrm { ACD }$\\
$\Rightarrow \quad \frac { 1 } { 2 } \mathrm { bc } \sin \mathrm { A } = \frac { 1 } { 2 } \mathrm { cAD } \sin \frac { \mathrm { A } } { 2 } + \frac { 1 } { 2 } \mathrm {~b} \times \mathrm { AD } \sin \frac { \mathrm { A } } { 2 }$\\
$\Rightarrow \quad \mathrm { AD } = \frac { 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \cos \frac { \mathrm { A } } { 2 }$\\
Again $\mathrm { AE } = \mathrm { AD } \sec \frac { \mathrm { A } } { 2 }$\\
$= \frac { 2 b c } { b + c } \Rightarrow A E$ is HM of $b$ and $c$.\\
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$\mathrm { EF } = \mathrm { ED } + \mathrm { DF } = 2 \mathrm { DE } = 2 \times \mathrm { AD } \tan \frac { \mathrm { A } } { 2 } = \frac { 2 \times 2 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \times \cos \frac { \mathrm { A } } { 2 } \times \tan \frac { \mathrm { A } } { 2 }$\\
$= \frac { 4 \mathrm { bc } } { \mathrm { b } + \mathrm { c } } \sin \frac { \mathrm { A } } { 2 }$\\
As $\mathrm { AD } \perp \mathrm { EF }$ and $\mathrm { DE } = \mathrm { DF }$ and AD is bisector ⇒ AEF is isosceles.\\
Hence A, B, C and D are correct answers.\\