[tex3]\mathsf{(\overline{OA})^2+(\overline{AT})^2+(\overline{TB})^2+(\overline{BO})^2=(\overline{AB})^2+(\overline{OT})^2+4\cdot(\overline{MN})^2 \ \text{Por Teo. Pitágoras }(\overline{TB})^2+(\overline{BO})^2=(\overline{OT})^2:\\
(\overline{OA})^2+(\overline{AT})^2+\cancel{(\overline{OT})^2}=(\overline{AB})^2+\cancel{(\overline{OT})^2}+4\cdot(\overline{MN})^2\\
(\overline{OA})^2+(\overline{AT})^2=(\overline{AB})^2+4\cdot(\overline{MN})^2 \ \text{ Foi dado que }(\overline{AO})^2+(\overline{AT})^2=16+(\overline{AB})^2:\\
16+(\overline{AB})^2=(\overline{AB})^2+4\cdot(\overline{MN})^2\\
4\cdot(\overline{MN})^2=16\\
(\overline{MN})^2=4\\
\boxed{\boxed{\color{red}\overline{MN}=2}}}[/tex3]
(Solução:rodBR -
viewtopic.php?t=89151)