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If median of the $\Delta \mathrm{ABC}$ through $\mathrm{A}$ is perpendicular to $\mathrm{BC}$, then which one of the following is correct?
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Verified Answer
The correct answer is:
$\tan B-\tan C=0$
In the given $\Delta \mathrm{ABC}$ let $\mathrm{BC}=\mathrm{a}$
$\therefore \quad \mathrm{BD}=\mathrm{CD}=\frac{\mathrm{a}}{2}$
In $\Delta \mathrm{ADB}$,
$\tan B=\frac{A D}{B D}=\frac{A D}{a / 2}$
$\Rightarrow \quad \tan B=\frac{2 \mathrm{AD}}{\mathrm{a}}$ ...(1)
In $\Delta \mathrm{ADC}$
$\tan \mathrm{C}=\frac{\mathrm{AD}}{\mathrm{CD}}=\frac{\mathrm{AD}}{\mathrm{a} / 2}$
$\tan C=\frac{2 A D}{a}$ ...(2)
From eqs. (1) and (2), we get
$\tan B=\tan C$
$\Rightarrow \tan B-\tan C=0$
$\therefore \quad \mathrm{BD}=\mathrm{CD}=\frac{\mathrm{a}}{2}$
In $\Delta \mathrm{ADB}$,
$\tan B=\frac{A D}{B D}=\frac{A D}{a / 2}$
$\Rightarrow \quad \tan B=\frac{2 \mathrm{AD}}{\mathrm{a}}$ ...(1)
In $\Delta \mathrm{ADC}$
$\tan \mathrm{C}=\frac{\mathrm{AD}}{\mathrm{CD}}=\frac{\mathrm{AD}}{\mathrm{a} / 2}$
$\tan C=\frac{2 A D}{a}$ ...(2)
From eqs. (1) and (2), we get
$\tan B=\tan C$
$\Rightarrow \tan B-\tan C=0$
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