a)∡EAF=60⇒∡FAD=30⇒ΔFAD teorema ∡de 30 ca DF=AF/2=6/2=3
ΔADF⇒teorema lui Pitagora AD∧2+DF∧2=AF∧2
AD=√36-9=√27=3√3
ΔABC⇒teorema lui Pitagora AB∧2+BC∧2=AC∧2
AC=√27+81=√108=6√3
b)daca DF=3⇒FC=6
cum ∡AFD=60, ∡AFE=60 ⇒∡CFE=60
FE=FC=6 atunci ΔFEC echilateral⇒EC=6
cum AF=FC=CE=AE=6
AE II FC obtinem ca AFCE romb
rombul are proprietatea ca diagonalele sunt perpendiculare deci AC⊥FE