Câu a : Theo tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{AB}{2}=\dfrac{AC}{3}=\dfrac{\sqrt{AB^2+AC^2}}{\sqrt{2^2+3^2}}=\dfrac{BC}{\sqrt{13}}=\dfrac{12}{\sqrt{13}}\)
\(\left\{{}\begin{matrix}\dfrac{AB}{2}=\dfrac{12}{\sqrt{13}}\Rightarrow AB=\dfrac{24}{\sqrt{13}}cm\\\dfrac{AC}{3}=\dfrac{12}{\sqrt{13}}\Rightarrow AC=\dfrac{36}{\sqrt{13}}cm\end{matrix}\right.\)
Câu b : Theo hệ thức lượng cho tam giác ABC ta có :
\(AI.BC=AB.AC\Rightarrow AI=\dfrac{AB.AC}{BC}=\dfrac{\dfrac{24}{\sqrt{13}}.\dfrac{36}{\sqrt{13}}}{12}=\dfrac{72}{13}cm\)
\(AB^2=AI.BC\Rightarrow AI=\dfrac{AB^2}{BC}=\dfrac{\left(\dfrac{24}{\sqrt{13}}\right)^2}{12}=\dfrac{48}{13}cm\)
