Câu3: Theo định lý Py-ta-go: \(BC=\sqrt{6^2+8^2}=\sqrt{100}=10\left(cm\right)\)
Xét ∆ABD và ∆CBA:
\(\widehat{B}\) : Góc chung
\(\widehat{ADB}=\widehat{CAB}=90^o\)
=> ∆ABD~∆CBA (g.g)
=> \(\dfrac{AD}{AC}=\dfrac{AB}{BC}\Rightarrow AD=\dfrac{AB.AC}{BC}=\dfrac{6.8}{10}=4,8\left(cm\right)\)
b) Theo câu a: ∆ABD~∆CBA
=> \(\dfrac{AB}{BD}=\dfrac{BC}{AB}\Rightarrow AB^2=BD.BC\)