a) Xét \(\Delta HBA,\Delta ABC\) có :
\(\left\{{}\begin{matrix}\widehat{B}:chung\\\widehat{AHB}=\widehat{CAB}\left(=90^o\right)\end{matrix}\right.\)
=> \(\Delta HBA\sim\Delta ABC\left(g.g\right)\) (1)
Xét \(\Delta HAC,\Delta ABC\) có :
\(\left\{{}\begin{matrix}\widehat{C}:Chung\\\widehat{AHC}=\widehat{BAC}\left(=90^o\right)\end{matrix}\right.\)
=> \(\Delta HAC\sim\Delta ABC\left(g.g\right)\) (2)
Từ (1) và (2) suy ra : \(\Delta HBA\sim\Delta HAC\left(\sim\Delta ABC\right)\)
b) Xét \(\Delta ABC\) có : \(\widehat{BAC}=90^o\) (gt)
\(\Rightarrow BC^2=AB^2+AC^2\) (Định lí Pitago)
\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{10^2+15^2}\approx18,03cm\)
Từ \(\Delta HBA\sim\Delta ABC\left(cmt\right)\) ta có :
\(\dfrac{AH}{AB}=\dfrac{AB}{BC}\)
\(\Rightarrow\dfrac{AH}{10}=\dfrac{10}{18,03}\)
\(\Rightarrow AH=\dfrac{10^2}{18,03}\approx5,55cm\)
Xét \(\Delta ABH\) có : \(\widehat{AHB}=90^o\left(gt\right)\)
\(\Rightarrow AB^2=HB^2+AH^2\)(Định lí Pitago)
\(\Rightarrow HB=\sqrt{AB^2-AH^2}=\sqrt{10^2-5,55^2}\approx8,32cm\)
\(\Rightarrow HC=BC-HB=18,03-8,32=9,71cm\)
c) \(S_{\Delta ABC}=\dfrac{AB.AC}{2}=\dfrac{10.15}{2}=75\left(cm^2\right)\)
d) Ta có : \(\Delta HBA\sim\Delta HAC\left(cmt\right)\)
Suy ra: \(\dfrac{HB}{AH}=\dfrac{AH}{HC}\)
\(\Rightarrow AH^2=HB.HC\)
