Question

In a triangle $ABC$, if the median $AD$ through $A$ is perpendicular to $AB$, then $AD$ is equal to:

• A. $c\tan B$
• B. $-c\tan A$
• C. $2c\tan A$
• D. $-\frac{c}{2}\tan A$

Answer 1

Answer: (A), (D)

For $△ACD$, use sine rule.

$\frac{\frac{a}{2}}{sin(A-90)}=\frac{AD}{\sin C}$

But, $\angle C=180-(A+B)$

and from $△ABD,AD=\frac{a}{2}sinB$

so, $\frac{\frac{a}{2}}{\sin (A-90)}=\frac{\frac{a}{2}\sin B}{\sin \{180-(A+B\left)\right\}}$

$⟹\frac{sinB}{\sin \left(\right(A+B)}=-\frac{1}{cosA}$

$⟹sinBcosA=-[sinAcosB+cosAsinB]$

$⟹tanA+2tanB=0⟶1$

From $△ABD,tanB=\frac{AD}{AB}$

$⟹AD=c\times tanB$

from $1$ no equation, $tanB=\frac{1}{2}tanA$

So, $AD=-\frac{c}{2}tanA$.