SA||OM (Om linie mijlocie in tr SAC), OM⊂(MBD) deci SA||(MBD)
b) SA=SC=l
AC=l√2 (diag de patrat)
Cu Rec TEo Pitagora rezulta imediat ΔSAC dreptunghic in S⇔SA⊥SC
aS Simple aS that!!
