Proof and guess are as follows (copy it if you don't understand it. . . . . . Don't want to explain)
Solution: (1) df = ef.
(2) conjecture: df = Fe.
It is proved that if point D is DG⊥AB in G, then ∠DGB=90 degrees.
DA = DB, ∠ADB=60 degrees.
∴AG=BG, △DBA is an equilateral triangle.
∴DB=BA.
∠∠ACB = 90 degrees ∠ABC=30 degrees,
∴AC=？ 1/2AB=BG。
∴△DBG≌△BAC.
∴DG=BC.
BE = EC，∠BEC=60，
△ EBC is an equilateral triangle.
∴bc=be∠CBE = 60 degrees.
∴DG=BE,∠ABE=∠ABC+∠CBE=90。
∠∠DFG =∠EFB∠DGF =∠EBF，
∴△DFG≌△EFB.
∴DF=EF.
(3) conjecture: df = Fe.
Method 1: If the intersection D is DH⊥AB in H, and HC, HE and he intersect with CB in K, then ∠DHB=90 degrees.
DA = DB，
∴AH=BH,∠ 1=∠HDB.
∫∠ACB = 90 degrees,
∴HC=HB.
EB = EC，HE=HE，
∴△HBE≌△HCE.
∴∠2=∠3,∠4=∠BEH.
∴HK⊥BC.
∴∠BKE=90。
∠∠ ADB =∠BEC=2∠ABC,
∴∠HDB=∠BEH=∠ABC.
∴∠DBC=∠DBH+∠ABC=∠DBH+∠HDB=90，
∠EBH=∠EBK+∠ABC=∠EBK+∠BEK=90。
∴DB∥HE,DH∥BE.
∴ Quadrilateral DHEB is a parallelogram.
∴DF=EF.
Method 2: If DH⊥AB is in H, EK⊥BC is in K and connected to HK, then
∠DHB=∠EKB=90 degrees.
∫∠ACB = 90 degrees,
∴EK∥AC.
DA = DB，EB=EC，
∴AH=BH,∠ 1=∠HDB，
CK=BK，∠2=∠BEK。
∴HK∥AC.
Point h, point k and point e are on the same straight line.
Same method 1.
2:( 1). Make it B'G‖BC, pass through the CD extension line to G. Then to B'G⊥B'C'.
∵AC⊥AC',AC=AC'
,∴∠ACC'=∠AC'C=45，∠B'C'G= 180 -90 -45 =45，
B'G⊥B'C'，
∴∠B'GC'=45，∠B'GC'=∠B'C'G，B'G=B'C'=BC，
So in △B'DG and △BCD,
Such as B'G‖BC, ∠B'GD=∠BCD, ∠B'DG=∠BDC, B'G=BC,
Therefore △ b' DG △ BCD, B'D=BD,
Because < cab+< c' ab = 90,
So ∠ b 'ac'+∠ c 'ad = 90,
So △AB'B is isosceles RT△,
D is that midpoint of the hypotenuse BB',
AD= 1/2BB '。
(2) If Rt△ABC rotates counterclockwise at any angle around point A, other conditions remain unchanged, and the conclusion (1) cannot be drawn.
Because △ABB' is not RT△, and the midline AD≠ 1/2BB'.
3:( 1)CK = EK；
Proof: BC = DE, AC=BE, ∠ ABC = ∠ BDE = 90,
∴△ABC≌△BDE，
∴ab=bd； ( 1)
∫M and n are the midpoint of AB and BD respectively, AB=2BC,
∴bm=am=bc= 1/2ab= 1/2bd=dn=bn，
∴∠BMN=∠BNM=∠DNE=∠BMC=45，
∴∠CMN=∠MNE=90，
Connect CM, EN,
Then △ BCM △ den
∴CM=NE, and ∞∠ckm =∠ekn,
∴△CMK≌△ENK，
∴ck=ek；
(2)CK = EK；
The straight line MK passes through the vertical sections of C and E, and the vertical feet are P and Q respectively.
From (1), we know △ ABC △ BDE, △ BCM △ den.
∴BM=BN,CM=NE,∠DNE=∠CMB，
∴∠BNM=∠BMN，
∴ 180-∠bnm-∠dne = 180-∠BmN-∠CMB，
Namely ∠CMP=∠ENQ,
And ∵∠ CPM = ∠ nqe = 90, CM=EN,
∴△CMP≌△ENQ，
∴PC=QE，
∠∠CPQ =∠EQP = 90，∠EKQ=∠CKP，
∴△CPK≌△EQK，
∴ck=ke；
(3) As shown in the figure, △ ABC △ BDE, M and N are the midpoints of AB and DB respectively, and the straight line MN and CE intersect at K. 。
Conclusion: CK = ek.
4. Point D is taken as the vertical line of BD, and B'D=BD is taken as the vertical line.
Make C'D perpendicular to CD when crossing point d, so C'D=CD.
Connecting bottom connection
Triangle DB'C' is the triangle to be made.
5. Connect AM and MC, and extend AM to point F, so that AM=MF,
Connect CF and DF, and extend the intersection of AO and DF at g point.
Because m is the midpoint of BD,
So AM=MFAM=MF, BM=DM, ∠AMB=∠DMC
So △ABM=△DMF,
So ∠ABM=∠MDF, AB=DF.
∫△ABO and △CDO are isosceles triangles, ∠ Bao = ∠ DCO = 90,
So AB=AO, CO=CD.
So AO=DF.
Because ∠ABM=∠MDF,
So ab sigma df,
So ∠ g = ∠ ADC = 90.
So ∠ CDF = 180-∠ COG = ∠ AOC. ∠CDF=∠AOC，CO=CD，AO=DF .
So δAOC =δCFD. (SAS)
So ∠ACO=∠DCF, AC = CF
So AM=CM and AM is perpendicular to CM.
So △AMC is an isosceles right triangle,
And ∵MN⊥AC,
∴2NM=AC (isosceles triangle connected by three lines)
I couldn't find your picture, so I read the original picture online. This is a new set of questions for 1 1 year. 1 1 month. The answer is my own, just compare it with what I know online. I hope it helps you? If you can't see the picture clearly, ask me, send me an email and I'll send you a big picture.