[tex]AB= \sqrt{(5-2)^{2}+(6-6)^{2} } \\
AB= \sqrt{9+0}
\\ AB= \sqrt{9}=3u
[/tex]
[tex]BC= \sqrt{(2-5)^{2}+(6-2)^{2} } \\
BC= \sqrt{(-3)^{2}+4^{2} } \\
BC= \sqrt{9+16} = \sqrt{25} =5[/tex]
[tex]AC= \sqrt{(5-5)^{2}+(6-2)^{2} } \\
AC= \sqrt{0+16}
AC= \sqrt{16}=4u
[/tex]
BC²=AB²+AC²
25=16+9
25=25(A)⇒ΔABC dreptunghic,m(<A)=90 grade
