In △ Abe and △CBF,

AB=BC∠A=∠C=90 AE=CF？ ,

∴△abe≌△cbf(sas)；

∴∠abe=∠cbf,be=bf；

∠∠ABC = 120，∠MBN=60，

∴∠ABE=∠CBF=30，

∴ae= 12be,cf= 12bf；

∫∠MBN = 60，BE=BF，

△ BEF is an equilateral triangle;

∴ae+cf= 12be+ 12bf=be=ef；

Figure 2 holds, but Figure 3 does not.

Prove figure 2.

Extend DC to point k, make CK=AE, connect BK,

At BAE Systems and BCK,

AB=CB∠A=∠BCK=90 AE=CK？

Then △ BAE △ BCK

∴BE=BK,∠ABE=∠KBC，

∠∠FBE = 60，∠ABC= 120，

∴∠FBC+∠ABE=60，

∴∠FBC+∠KBC=60，

∴∠KBF=∠FBE=60，

At △KBF and △EBF,

BK=BE∠KBF=∠EBFBF=BF？

∴△KBF≌△EBF，

∴KF=EF，

∴KC+CF=EF，

That is AE+cf = ef.

Fig. 3 does not hold,

The relationship among AE, CF and EF is AE -CF=EF.

Hope to adopt