a) \(\text{Xét }\Delta\text{ vuông }HBA\text{ và }\Delta\text{ vuông }ABC,\text{có: }\)
\(\widehat{BHA}=\widehat{BAC}=90^o\)
\(\widehat{ABC}\text{ chung }\)
\(\Rightarrow\Delta\text{ vuông }HBA\text{ đồng dạng }\Delta\text{ vuông }ABC\)
\(\Rightarrow\frac{HB}{AB}=\frac{HA}{AC}=\frac{AB}{BC}\)
\(\frac{BH}{AB}=\frac{AB}{BC}\Leftrightarrow BC.HB=AB^2\)
b) cm tương tự câu a:
t/g vuông BAC đồng dạng t/g vuông AHC
\(\Rightarrow\frac{AB}{AH}=\frac{BC}{AC}=\frac{AC}{HC}\)
\(\frac{BC}{AC}=\frac{AC}{HC}\Leftrightarrow AC^2=BC.HC\)
c) \(\frac{HA}{AC}=\frac{AB}{BC}\left(\text{câu a}\right)\)
\(\Leftrightarrow HA.BC=AC.AB\)
d) \(\left\{{}\begin{matrix}\frac{AH}{AC}=\frac{HB}{AB}\\\frac{AB}{AH}=\frac{AC}{HC}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}AH=\frac{HB.AC}{AB}\\AH=\frac{AB.HC}{AC}\end{matrix}\right.\)
\(\Leftrightarrow AH^2=\frac{HB.AC}{AB}\cdot\frac{AB.HC}{AC}=HB.HC\)
