a) Xét \(\Delta HAC\) và \(\Delta ABC\) có:
\(\left\{{}\begin{matrix}\widehat{C}:chung\\A\widehat{H}C=B\widehat{A}C=90\end{matrix}\right.\)\(\Rightarrow\Delta HAC\wr\Delta ABC\left(g.g\right)\)
b) Vì BD là phân giác góc B nên:
\(\Rightarrow\dfrac{AD}{CD}=\dfrac{AB}{BC}\) (1)
Ta có: \(\left\{{}\begin{matrix}\widehat{B}:chung\\A\widehat{H}B=C\widehat{A}B=90\end{matrix}\right.\)\(\Rightarrow\Delta AHB\wr\Delta CAB\)
\(\Rightarrow\dfrac{AB}{BC}=\dfrac{AH}{AC}\) (2)
Từ (1);(2)\(\Rightarrow\dfrac{AD}{CD}=\dfrac{AH}{AC}\Leftrightarrow AC\cdot AD=AH\cdot CD\)
c) Ta có: \(H\widehat{C}A=H\widehat{A}B\left(\Delta AHB\wr\Delta CAB\right)\)
Và \(A\widehat{H}C=B\widehat{H}A=90\)
Do đó \(\Delta HAC\wr\Delta HBA\left(g.g\right)\)
\(\Rightarrow k=\dfrac{HA}{HB}=\dfrac{AC}{AB}=\dfrac{8}{4}=2\)
\(\Rightarrow\dfrac{S_{\Delta HAC}}{S_{\Delta HBA}}=k^2=2^2=4\)
Vậy..................
