a) Xét ΔHBA và ΔABC có:
\(\widehat{BAC}=\widehat{BHA}\) \(=90^0\)
\(\widehat{ABC}:chung\)
=> ΔHBA∼ΔABC (g.g)
b) Vì \(\widehat{C}\) \(=\frac{1}{2}\)\(\widehat{ABC}\) (mà \(\widehat{EBA}=\widehat{EBC}\) vì BE là p/g \(\widehat{ABC}\))
=> \(\widehat{C}=\widehat{ABE}=\widehat{EBC}\) \(=\frac{1}{2}\)\(\widehat{ABC}\)
Xét ΔABE và ΔACB có:
Â: chung
\(\widehat{ABE}=\widehat{C} (cmtrn)\)
=> ΔABE∼ΔACB (g.g)
=> \(\frac{AB}{AC}=\frac{AE}{AB}\Leftrightarrow AB^2=AE.AC\left(đpcm\right)\)
c) Theo câu a) ta có: ΔHBA∼ΔABC
\(\Rightarrow\frac{BH}{AB}=\frac{AB}{BC}\Leftrightarrow BH.BC=AB^2\)
\(\Leftrightarrow6BH=3^2\Leftrightarrow BH=\frac{3^2}{6}=1,5\left(cm\right)\)
Xét ΔBHD và ΔBAE có:
\(\widehat{BAE}=\widehat{BHD}\) \(=90^0\)
\(\widehat{EBA}=\widehat{DBH}\) (cmtrn)
=> ΔBHD∼ΔBAE (g.g)
=> \(\frac{S_{BHD}}{S_{BAE}}=\left(\frac{BH}{AB}\right)^2=\left(\frac{1,5}{3}\right)^2=\frac{1}{4}\)
