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