a) Xét \(\Delta HBA\) và \(\Delta HAC\) có:
\(\widehat{AHB}=\widehat{CHA}=90^0\)
\(\widehat{HBA}=\widehat{HAC}\) do cùng phụ với góc HAB
suy ra: \(\Delta HBA~\Delta HAC\)
b) Xét \(\Delta ABC\)và \(\Delta HBA\) có:
\(\widehat{BAC}=\widehat{BHA}=90^0\)
\(\widehat{B}\) CHUNG
suy ra: \(\Delta ABC~\Delta HBA\)
\(\Rightarrow\)\(\frac{AB}{HB}=\frac{BC}{AB}\)
\(\Leftrightarrow\)\(AB^2=BH.BC\) (ĐPCM)
c) \(\Delta HBA~\Delta HAC\) \(\Rightarrow\) \(\frac{S_{HBA}}{S_{HAC}}=\left(\frac{AB}{AC}\right)^2=\left(\frac{3}{4}\right)^2=\frac{9}{16}\)
HAY \(\frac{S_{HBA}}{32}=\frac{9}{16}\) \(\Rightarrow\)\(S_{HBA}=\frac{32.9}{16}=18\)
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