Notam triunghiul isoscel ABC, cu AB=AC si baza BC=4√2.
Notam M mijlocul lui AC.
Inseamna ca mediana BM are lungimea 5.
Conform teoremei medianei:
[tex]BM^2=\frac{2(AB^2+BC^2)-AC^2}{4}[/tex]
Dar [tex]AB=AC[/tex], deci
[tex]BM^2=\frac{2AB^2+2BC^2-AB^2}{4}=\frac{AB^2+2BC^2}{4}[/tex]
Cunoastem [tex]BM[/tex] si [tex]BC[/tex], deci
[tex]5^2=\frac{AB^2+2\times (4\sqrt2)^2}{4} \implies 4\times 25=AB^2+64\implies AB^2=100-64=36[/tex]
[tex]\implies AB=6[/tex]
Deci laturile laterale au lungimea de 6 cm.
