a) \(\Delta HBA\) và \(\Delta ABC\) ta có:
\(\widehat{B}\) chung
\(\widehat{BHA}=\widehat{BAC}=90^o\)
\(\Rightarrow\Delta HBA\sim\Delta ABC\left(g.g\right)\)
\(\Rightarrow\dfrac{HB}{AB}=\dfrac{BA}{BC}=\dfrac{HA}{AC}\)
b)
\(\Delta ABC\) vuông tại \(A\) ta có:
\(BC^2=AB^2+AC^2\) (định lí Pi ta go)
\(\Leftrightarrow BC^2=15^2+20^2=625\)
\(\Rightarrow BC=\sqrt{625}=25\left(cm\right)\)
\(\dfrac{BA}{BC}=\dfrac{HA}{AC}\left(cmt\right)\Leftrightarrow\dfrac{15}{25}=\dfrac{HA}{20}\Rightarrow HA=\dfrac{15\times20}{25}=12\left(cm\right)\)
\(\dfrac{HB}{AB}=\dfrac{BA}{BC}\left(cmt\right)\Leftrightarrow\dfrac{HB}{15}=\dfrac{15}{25}\Rightarrow HB=\dfrac{15\times15}{25}=9\left(cm\right)\)
\(CH=BC-BH=25-9=16\left(cm\right)\)
Vậy \(BC=25\left(cm\right);\) \(AH=12\left(cm\right);\) \(BH=9\left(cm\right);\) \(CH=16\left(cm\right).\)
