Đáp án+Giải thích các bước giải:

$a)$ Xét $\Delta BHA $ và $\Delta BAC:$

$\widehat{B}:$ chung

$\widehat{BHA}=\widehat{BAC}=90^\circ\\ \Rightarrow \Delta BHA \backsim \Delta BAC (g.g)(1)$

Xét $\Delta AHC$ và $\Delta BAC:$

$\widehat{C}:$ chung

$\widehat{AHC}=\widehat{BAC}=90^\circ\\ \Rightarrow \Delta AHC \backsim \Delta BAC (g.g)(2)\\ (1)(2) \Rightarrow \Delta BHA \backsim \Delta AHC\\ b) \Delta BHA \backsim \Delta AHC\\ \Rightarrow \dfrac{AH}{HC}=\dfrac{BH}{AH}\Rightarrow AH^2=BH.CH\\ \Delta BHA \backsim \Delta BAC\\ \Rightarrow \dfrac{AB}{BC}=\dfrac{BH}{AB} \Rightarrow  AB^2=BH.BC\\ \Delta AHC \backsim \Delta BAC\\ \Rightarrow \dfrac{AC}{BC}=\dfrac{HC}{AC} \Rightarrow AC^2=CH.BC\\ \left.\begin{array}{l} S_{ABC}=\dfrac{1}{2}AH.BC\\S_{ABC}=\dfrac{1}{2}AB.AC\end{array} \right\}\Rightarrow AH.BC=AB.AC.$