The first question is simple: connecting OE, OF, we can know that OE=BE, OF=FC, △OEF is an equilateral triangle, so BE=EF=FC.

The second question: it is also well proved. Because ED=EB and DF=FC.

So, get a license.

Both of these problems make use of the properties of angular bisector. You came out as soon as you painted.

The following is a detailed proof.

( 1)

∫△ABC is an equilateral triangle

∴∠ABC=∠ACB=60

Ob and OC are angular bisectors.

∴∠OBC=∠OCB=30

E is on the vertical line of OB.

∴EB=EO

∴∠OEF=2∠OBE=60

Similarly: FO=FC, ∠ ofe = 60.

∴△OEF is an equilateral triangle

∴OF=EF

∴BE=EF=FC

(2)

EF communication

∴∠EDB=∠DBC

∫BD equal division ∠ABC

∴∠EBD=∠CBD

∴∠EBD=∠EDB

∴ED=EB

Similarly: DF=FC

∴EF=ED+FD=BE+CF