in Δ ABC echilateral, construim CM⊥ AB⇒ CM-mediana si inaltime
in Δ ABD echilateral, construim DM⊥ AB⇒ DM-mediana si inaltime
AB=AC=BC=BD=AD⇒ΔABC≡ΔABD⇒ CM=DM =l√3/2=2√3/2=√3
(ABC) n (ABD)=AB
CM⊥AB, CM⊂ (ABC)
DM⊥AB, DM⊂ (ABD) ⇒ ∡(DM;CM) =90°
⇒ΔCMD dreptunghic isoscel
CD = ipotenuza = cateta√2 = √3*√2=√6