There is a clerical error in the problem, which should be: the parallel line connecting CE and DE certificate: BCD is made by point E, and the extension line of CA intersects with F because EBD, ∠BEF=∠EBD=60 degrees; Because ∠EAF=∠BAC=60 degrees, the triangle AEF is an equilateral triangle; So: AE=AF=BD. In triangle BDE and triangle CEF, ∠EBD=60 degrees =∠CFE, BE=BA+AE=CA+AF=CF, AE=AF, so triangle BDE and triangle CEF are identical, so: EC=ED. Certificate of completion