[tex]\it AB\in\mathbb{Q} \Rightarrow AB^2\in\mathbb{Q}\ \ \ \ (1)\\ \\ AC\in\mathbb{Q}\Rightarrow AC^2\in\mathbb{Q}\ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow AB^2+AC^2\in\mathbb{Q} \stackrel{T.Pitagora}{\Longrightarrow}\ BC^2\in\mathbb{Q}\ \ \ \ (3)\\ \\ AD=\dfrac{AB\cdot AC}{BC} \Rightarrow AD^2=\dfrac{AB^2\cdot AC^2}{BC^2}\in\mathbb{Q}\ \ \ \ (4)[/tex]
[tex]\it AD^2=\dfrac{AB^2\cdot AC^2}{BC^2} \Rightarrow AB^2\cdot AC^2=BC^2\cdot AD^2\ \ \ \ \ (5)\\ \\ \\ AB^4+AD^4+AC^4 =(AB^2+AC^2)^2-2AB^2\cdot AC^2+AD^4\ \stackrel{(5)}{=}\ \\ \\ =BC^4-2BC^2\cdot AD^2+AD^4 =(BC^2-AD^2)^2[/tex]
Folosind ultima relație, rezultă:
[tex]\it \sqrt{AB^4+AD^4+AC^4}=\sqrt{(BC^2-AD^2)^2} =BC^2-AD^2\in\mathbb{Q}[/tex]
