Avem Δ <A=90, AD_|_BC (vezi desen atasat)
AD = 3√5cm
BC = 18 cm
Ducem mediana AE =BE=EC=BC:2=9
Consideram ΔADE care este dreptunghic, <ADE=90
unde avem:
AD²+DE²=AE²
(3√5)²+DE²=9²
DE²=81-45=36
DE=6
BD=BE-DE=9-6
BD=3
DC=BC-BD=18-3
DC=15
Din teorema catetei avem:
AC²=CD*BC=15*18=270
AC=3√10
AB²=BD*BC=3*18=54
AB=3√6