And ∠ ABC = 120, ∠ EBF = 60, ∠ Abe +∠CBF+∠EBF=∠ABC, we can see ∠ Abe = ∠ CBF = (6544 .
△BAE is a right triangle with a vertex angle of 30 degrees, so AE=BE/2. Similarly CF=BF/2.
From the congruence of triangle, BE=BF, EBF angle is 60 degrees, and it can be known that △BEF is an equilateral triangle. BE=BF=EF。
Therefore EF=AE+BF.
(2) If AE is not equal to CF, Pythagorean theorem
BA^2+AE^2=BE^2
BC^2+CF^2=BF^2
According to the formula of triangle side length, for triangle BEF,
EF^2=BE^2+BF^2-2*BE*BF*cos∠EBF
=ba^2+ae^2+bc^2+cf^2-2*√(ba^2+ae^2)*√(bc^2+cf^2)*√3/2
=2ab^2+ae^2cf^2-√[3*(ab^2+ae^2)(ab^2+cf^2)]( 1)
There is another formula, that is
tan∠ABE = AE/AB； tan∠CBF = CF/BC； ∠ Abe +∠CBF=60
It is obtained by the formula tan (a+b) = (tana+tanb)/(1-tana * tanb).
tan 60 =(ae/ab+cf/ab)/( 1-ae*cf/ab^2)
√3=[(ae+cf)/ab]/[(ab^2-ae*cf)/ab^2]
√3(AB^2-AE*CF)=(AE+CF)*AB
simplify
√3AB^2-(AE+CF)*AB-AE*CF=0
The relationship between AB and AE and CF can be obtained, and the general formulas of EF, AE and CF can be obtained by substituting (1) to eliminate AB.