daca CDFE este patrat=> CD=EF==DF=CE=3
=> T Pitagora
[tex] {ec}^{2} + {eb}^{2} = {cb }^{2} = > \\ {3}^{2} + {eb}^{2} = {5}^{2} \\ {eb}^{2} = 25 - 9 = 16 \\ eb = \sqrt{16} = 4[/tex]
AF=EB=4
[tex]mn = \frac{3 +11 }{2} = \frac{14}{2} = 7 \\ [/tex]
AB=2×EB+FE= 2×4+3= 8+3=11
