Ta có:
$\widehat{C} = \widehat{AHE}$ (cùng phụ $\widehat{HAE}$)
$\widehat{AHE} = \widehat{DHB}$ (đối đỉnh)
$\Rightarrow \widehat{C} = \widehat{DHB}$
$\Rightarrow tanC = tan\widehat{DHB} = \dfrac{BD}{HD}$
Ta được:
$tanB.tanC = \dfrac{AD}{BD}.\dfrac{BD}{HD} = \dfrac{AD}{HD}$
Ta lại có:
$\dfrac{HD}{HA} = \dfrac{1}{2}$ $(gt)$
$\Rightarrow HD = \dfrac{1}{2}HA$
$\Rightarrow HD = \dfrac{1}{3}AD$
Do đó: $\dfrac{AD}{HD} = 3$
$\Rightarrow tanB.tanC = 3$