DE//AB, deci BDEA trapez
mas ∡(BDE)=180°-mas∡(ABC)=180°-60°=120°
E mijloc AC, ED//AB , ED l.m . in tr ABC⇒ED=AB/2
ED//AB⇒(Teorema fundamentala a asemanarii) tr.EDC echilateral, ED=EC=AC/2=AB/2
⇒BD=ED⇔trBDCisoscel⇒mas∡(BED)=(180°-120°)/2=60°/2=30°
altfel
ED//AB, T.F.A RDC echilateral
⇒mas∡(DEC)=60°
AE=EC, BE mediana in ABC, ABC echilateral, BE inaltime⇔mas∡(BEC)=90°
mas∡(BED)=mas∡(BEC)-mas∡(DEC)= 90°-60°=30°
Obs.: o constructie corecta ajuta foarte mult