Let ABC be an acute-angled triangle and let D be the mid-point of...

In $∆ ABC, D$ be the mid-point of $BC and$AB = AD ⇒ \Delta ABD$is isosceles.</p><p>Let$AM$be the perpendicular drawn from$A$to$BC$.</p><p>$∴ AM$is perpendicular bisector of$BD$</p><p>$BM = MD$</p><p>$ln\Delta ABM, tanB = \dfrac{AM}{BM}$</p><p><img src="/api/proxy-image?url=https$ln\Delta AMC, tanC = \dfrac{AM}{CM}$</p><p>$∴ \dfrac{tanB}{tanC} = \dfrac{CM}{BM} = \dfrac{3BM}{BM} = 3 $\left[∵ CM = DM + CD\right]</p>$