Lời giải:
Ta có: $HA = HD =\dfrac12AD$
$\Leftrightarrow \dfrac{AD}{HD} = 2$
Xét $\triangle BHD$ và $\triangle ACD$ có:
$\begin{cases}\widehat{BDH} = \widehat{ADC} = 90^\circ\\\widehat{HBD} = \widehat{DAC}\quad \text{(cùng phụ $\widehat{C}$)}\end{cases}$
Do đó $\triangle BHD\backsim \triangle ACD\ (g.g)$
$\Rightarrow \dfrac{HD}{CD} = \dfrac{BD}{AD}$
Khi đó:
$\quad \tan B\cdot \tan C = \dfrac{AD}{BD}\cdot \dfrac{AD}{CD}$
$\Leftrightarrow \tan B\cdot \tan C = \dfrac{CD}{HD}\cdot \dfrac{AD}{CD}$
$\Leftrightarrow \tan B\cdot \tan C = \dfrac{AD}{HD}$
$\Leftrightarrow \tan B\cdot \tan C = 2$
