a. Xét \(\Delta BHA\left(\widehat{BHA}=90^o\right)\) có:
\(AB^2=BH^2+AH^2\) ( theo định lí Ta-let)
\(\Rightarrow\left(\sqrt{32}\right)^2=BH^2+4^2\)
\(\Rightarrow BH^2=16\)
\(\Rightarrow BH=4\left(cm\right)\)
Ta có: \(BC=BH+HC=4+4=8\left(cm\right)\)
Xét \(\Delta ABC\left(\widehat{BAC}=90^o\right)\) có:
\(BC^2=AB^2+AC^2\) ( theo định lí Ta-let)
\(\Rightarrow8^2=\left(\sqrt{32}\right)^2+AC^2\)
\(\Rightarrow AC^2=32\)
\(\Rightarrow AC=\sqrt{32}\left(cm\right)\)
\(\Rightarrow AB=AC\Rightarrow\Delta ABC\) vuông cân tại A.
b. Ta có: \(HC=HA\left(=4cm\right)\Rightarrow\Delta HAC\) vuông cân tại H
\(\Rightarrow\widehat{HAC}=\frac{180^o-90^o}{2}=45^o\)
