Special topic: computational problems; Query type.
Analysis: (1) Because △AOB is an isosceles right triangle and A (4 4,4) is AE⊥OB in E, the coordinates of point B can be obtained;
(2) Let AE⊥OB be in E, DF⊥OB be in F, and verify △ DFC △ CEA. Then, according to the equivalent transformation, it is proved that △AOB is an isosceles right triangle, and then the degree of △ ∠AOD can be obtained;
(3) The equation holds. It is easy to prove △EAN≌△AE⊥OB by intercepting AN=OF, even en. Then it is proved that △ nem △ FEM has AM-MF=OF according to the angle relation, and the equation is established. Solution: (1)
∫A(4，4)，
∴OE=4，
∵△AOB is an isosceles right triangle, while AE⊥OB,
∴OE=EB=4，
∴OB=8，
∴b(8,0)；
(2) Let AE⊥OB be in E, DF⊥OB be in F,
△ ACD is an isosceles right triangle,
∴AC=DC,∠ACD=90
That is, < ACF+< DCF = 90,
∫∠FDC+∠DCF = 90，
∴∠ACF=∠FDC，
∠∠DFC =∠AEC = 90，
∴△DFC≌△CEA，
∴EC=DF,FC=AE，
∫A(4，4)，
∴AE=OE=4，
∴FC=OE, that is, +EF=CE+EF,
∴OF=CE，
∴OF=DF，
∴∠DOF=45，
∫△AOB is an isosceles right triangle,
∴∠AOB=45，
∴∠aod=∠aob+∠dof=90；
Method 1: the extension line of CK⊥x axis intersecting OA is at K,
Then △OCK is an isosceles right triangle, OC=CK, ∠ k = 45,
∫△ACD is isosceles Rt△,
∴∠ACK=90 -∠OCA=∠DCO，AC=DC，
∴△ACK≌△DCO(SAS)，
∴∠DOC=∠K=45，
∴∠aod=∠aob+∠doc=90；
(3) (AM-MF)/OF= 1 is established for the following reasons:
Intercept AN=OF, even en on AM.
∫A(4，4)，
∴AE=OE=4，
∵∠ ean =∠ eof = 90，AN=OF，
∴△EAN≌△EOF(SAS)，
∴∠OEF=∠AEN,EF=EN，
And ∵△EGH are isosceles right triangles,
Geh = 45, that is, OEF+OEM = 45,
∴∠AEN+∠OEM=45
∫∠AEO = 90 degrees,
∴∠NEM=45 =∠FEM，
EM = EM once again,
∴△NEM≌△FEM(SAS)，
∴MN=MF，
∴AM-MF=AM-MN=AN，
∴AM-MF=OF，
Namely (am-MF)/of =1;
Method 2: intercept ON=AM, EN, MN,
Then △ EAM △ eon (SAS), EN=EM, ∠NEO=∠MEA,
That is, ∠NEF+∠FEO=∠MEA, and ∠ MEA+∠ MEO = 90,
∴∠ NEF+∠ FeO+∠ MEO = 90, while ∠ FeO+∠ MEO = 45,
∴∠nef=45 =∠mef,∴△nef≌△mef(sas),∴nf=mf，
∴AM=OF=OF+NF=OF+MF, that is, (am-MF)/of = 1.
(1) is the 2B point coordinate of your problem, and (2) is the AOD degree of your problem 1. Ask me if you don't understand.
I have done this problem, and there is another 1 answer.
Title: As shown in the figure, in the plane rectangular coordinate system, △AOB is an isosceles right triangle, A (4 4,4).
(1) Find the coordinates of point B;
(2) If C is the moving point on the positive semi-axis of X axis, take AC as the right angle to make an isosceles right angle △ACD, ∠ ACD = 90, or even OD, and find the degree of ∠AOD;
(3) When the intersection point A intersects with E as the vertical line of the Y axis, F as a point on the negative semi-axis of the X axis, G as a point on the extension line of EF, EG as a right-angled side as an isosceles Rt△EGH, and intersection point A as the vertical line of the X axis intersects with EH at M point or even FM point, does the equation = 1 hold? If yes, please prove it; If not, explain why.