Is it? +BC？ -CE？ = 2cos∠B X BE X BC

Push 2? +BC？ -39/2×2BC=cos60 = 1/2

The solution is BC=7.

Because the quadrilateral ABCD is a diamond, AB = BC = 7；;

2. Prove:

∫ Quadrilateral ABCD is a rhombus with four equal sides, ∠B=60?

∴⊿ABC, ⊿ACD are equilateral triangles, BC=AC, ∠B=∠CAD=60?

Intercept BG=AF on AB and connect CG and CF.

Then ⊿BCG≌⊿ACF(SAS).

∴CG=CF,∠BCG=∠ACF

∠∠BCG+∠ACG =∠ACB = 60？

∴∠GCF=∠ACF+∠ACG=60？

∴⊿GCF is an equilateral triangle

∴G point is on the middle vertical line of CF, and the point with the same distance at both ends of the line segment is on the middle vertical line of the line segment.

CE = EF

Point e is also on the middle vertical line of cf.

∴ Point G and point E coincide. There is only one intersection between the perpendicular line of CF and AB.

That is, BE=BG=AF.

∴BC=AB=AE+BE=AE+AF