Reason: ∵AB⊥AD, BC⊥CD, AB=BC, AE = CF.

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= 1/2 Yes

CF = 1/2BF；

∫∠MBN = 60，BE=BF，

△ BEF is an equilateral triangle;

∴ae+cf= 1/2be+ 1/2bf=be=ef；

(2) As shown in Figure 2, the conclusion in (1) holds.

Proof: extend FC to h, make CH=AE, connect BH,

∵AB⊥AD,BC⊥CD，

∴∠A=∠BCH=90，

In △BCH and △BAE

BC=AB

{∠BCH=∠A

CH=AE

∴△BCH≌△BAE(SAS)，

∴BH=BE,∠CBH=∠ABE，

∠∠ABC = 120，∠MBN=60，

∴∠ABE+∠CBF= 120 -60 =60，

∴∠HBC+∠CBF=60，

∴∠HBF=60 =∠MBN，

At △HBF and △EBF,

BH=BE

{∠HBF=∠EBF

BF=BF

∴△HBF≌△EBF(SAS)，

∴HF=EF，

∫HF = HC+CF = AE+CF，

∴EF=AE+CF.

The conclusion in figure 3 is not valid, and the quantitative relationship of line segments AE, CF and EF is AE=EF+CF,

Proof: intercept AQ=CF on AE and connect BQ.

∵AB⊥AD,BC⊥CD，

∴∠A=∠BCF=90，

At △BCF and △BAQ,

BC=AB

{∠BCF=∠A

CF=AQ

∴△BCF≌△BAQ(SAS)，

∴BF=BQ,∠CBF=∠ABQ，

∠∠MBN = 60 =∠CBF+∠CBE，

∴∠CBE+∠ABQ=60，

∫∠ABC = 120，

∴∠QBE= 120 -60 =60 =∠MBN，

At △FBE and △QBE,

BF=BQ

{∠FBE=∠QBE

Yes = yes

∴△FBE≌△QBE(SAS)，

∴EF=QE，

∫AE = QE+AQ = EF+CF，

∴AE=EF+CF，

That is, the conclusion in (1) is not valid, and the quantitative relationship between AE, CF and EF is AE = EF+CF. 。

See the figure for details.

/math/ques/detail/7ff 90 ace-d4e 1-47 a3-ba F3-aa8e 4408 e 99 1