Cu teorema lui Pitagora in triunghiul ABD avem BD^2=15^2+8^2=289, BD=17.
Avem asemanarea:
[tex]\triangle AMB \sim \triangle NMD(T.F.A.)\\
\frac{DN}{AB}=\frac{DM}{MB}\\
\frac{DN+AB}{AB}=\frac{DM+MB}{MB}\\
\frac{21}{15}=\frac{DB}{MB} \\
\frac{7}{5}=\frac{17}{MB}\\
MB=\frac{17\cdot 5}{7}\\
MB=\frac{85}{7}=12\frac{1}{7}
[/tex]
