AC=AB√2 , diagonala de patrat
VA≡AB≡VC
AC²=(AB√2 )²=2AB²
VA²+VC²=AB²+AB²=2AB²=AC²
conform Rec. Teo Pitagora, ΔVAC dreptunghic de ipotenuza AC⇔AV⊥CV, cerinta
VA≡AB≡VC=BC⇒ΔVBC echilateral
mas ∡(VB, AD)=mas ∡(VB, BC)=60°
mas ∡( VA, AC)=?
VA=VC (ipoteza)
VA⊥VC demonstratie, pct a)⇒ΔVAC dreptunghic isoscel⇒mas ∡( VA, AC)=45°
