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The first volume of the eighth grade math problem (related to triangles)
(1) solution: as shown in figure 1, AE+CF=EF,

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