Let $ABC$ be an acute-angled triangle and let $D$ be the mid-point of $BC$. If $AB=AD$, then $\frac{\tan B}{\tan C}$ equal

In  $∆ABC,D$ be the mid-point of $BC$ and $AB=AD\Rightarrow \mathrm{\Delta }ABD$ is isosceles.

Let $AM$ be the perpendicular drawn from $A$ to $BC$.

$\therefore AM$ is perpendicular bisector of $BD$

$BM=MD$

$\ln \mathrm{\Delta }ABM, \tan B=\frac{AM}{BM}$

$\ln \Delta AMC, \tan C=\frac{AM}{CM}$

$\therefore \frac{\tan B}{\tan C}=\frac{CM}{BM}=\frac{3BM}{BM}=3 \left[\because CM=DM+CD\right]$