Câu a : Theo tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{BD}{3}=\dfrac{CD}{4}=\dfrac{BD+CD}{3+4}=\dfrac{BC}{7}=\dfrac{10}{7}\)
\(\left\{{}\begin{matrix}\dfrac{BD}{3}=\dfrac{10}{7}\Rightarrow BD=\dfrac{30}{7}cm\\\dfrac{CD}{4}=\dfrac{10}{7}\Rightarrow CD=\dfrac{40}{7}cm\end{matrix}\right.\)
Theo tính chất đường phân giác ta có :
\(\dfrac{AB}{BD}=\dfrac{AC}{CD}=\dfrac{\sqrt{AB^2+AC^2}}{\sqrt{BD^2+CD^2}}=\dfrac{BC}{\sqrt{\dfrac{30^2}{7^2}+\dfrac{40^2}{7^2}}}=\dfrac{10}{\dfrac{50}{7}}=\dfrac{7}{5}\)
\(\left\{{}\begin{matrix}\dfrac{AB}{\dfrac{30}{7}}=\dfrac{7}{5}\Rightarrow AB=6cm\\\dfrac{AC}{\dfrac{40}{7}}=\dfrac{7}{5}\Rightarrow AC=8cm\end{matrix}\right.\)
Câu b : Theo hệ thức lượng cho tam giác ABC ta có :
\(\dfrac{1}{AH^2}=\dfrac{1}{AB^2}+\dfrac{1}{AC^2}\Rightarrow AH=\sqrt{\dfrac{1}{\dfrac{1}{AB^2}+\dfrac{1}{AC^2}}}=\sqrt{\dfrac{1}{\dfrac{1}{6^2}+\dfrac{1}{8^2}}}=4,8cm\)
