It is known that AB and CD are the two diameters of ⊙ o, and the chord CE ∥ ab. it is proved that ad = AE

It is known that AB and CD are the two diameters of ⊙ o, and the chord CE ∥ ab. it is proved that ad = AE

Proof: connect BC,
∵ AB and CD are the two diameters of ⊙ o, ∵ AOD = ∠ BOC,
ν arc BC = arc ad
∵CE∥AB，
ν arc BC = arc AE
The arc ad = arc AE
∴AD=AE．

As shown in the figure, ab = AC, ad = AE are known

It is proved that AF ⊥ BC is applied to F,
∵ AB = AC (known),
ν BF = CF (three wires in one),
And ∵ ad = AE (known),
/ / DF = ef (three wires in one),
ν bf-df = cf-ef, i.e. BD = Ce (property of the equation)

As shown in the figure, AB, CD are the diameters of ⊙ o, CE / / AB intersects with E and connects AD / AE. It is proved that ad = AE

Connect EO
∵AB||CE
∴∠ECD=∠AOD
∵ the circle angle of the arc ead is ∠ ECD, and the center angle of the arc is ∠ EOD
∴∠ECD=1/2∠EOD
∴∠EOA=∠AOD
The arc ad is equal to the arc AE
∴AD=AE

Known: as shown in the figure, e and F are points on BC, BF = CE, points a and D are on both sides of BC, and AE ∥ DF, AE = DF Confirmation: ab ∥ CD

Proof: ∵ AE ∥ DF,
∴∠AEB=∠DFC．
∵BF=CE，
∴BF+EF=CE+EF．
Be = CF
∵ in ᙽ Abe and ∵ DCF,
AE＝DF
∠AEB＝∠DFC
BE＝CF ，
∴△ABE≌△DCF（SAS）．
∴∠B=∠C．
∴AB∥CD．

It is known that: as shown in the figure, AB and CD intersect at O, AC ∥ BD, OC = OD, e, f are the two points on AB, and AE = BF

Proof: ∵ AC ∥ BD,
∴∠ACO=∠BDO，
In △ ACO and △ BDO,
∠ACO＝∠BDO
OC＝OD
∠AOC＝∠BOD ，
∴△ACO≌△BDO（ASA），
∴OA=OB，
∵AE=BF，
∴OE=OF，
In △ CoE and △ DOF,
OC＝OD
∠COE＝∠DOF
OE＝OF ，
∴△COE≌△DOF（SAS），
∴CE=DF．

As shown in the figure, ab = CD, AE ⊥ BC, DF ⊥ BC, CE = BF

Proof: AE ⊥ BC, DF ⊥ BC,
∴∠DFC=∠AEB=90°，
And ∵ CE = BF,
ν ce-ef = bf-ef, that is, CF = be,
∵AB=CD，
∴Rt△DFC≌Rt△AEB（HL），
∴AE=DF．

As shown in the figure, it is known that ab = CD, AE = BF, CE = DF, and it is proved that: ∠ e = ∠ F

Proof: ab = CD,
∴AB+CB=CD+CB，
That is, AC = dB,
In △ AEC and △ BFD,
AE＝BF
CE＝DF
AC＝BD ，
∴△AEC≌△BFD（SSS），
∴∠E=∠F．

It is known that in the parallelogram ABCD, e and F are two points on AB and CD respectively, and AE = CF, AF, DF intersect at point m, BF and CE intersect at point n Proving that the quadrilateral EMFN is a parallelogram

It is proved that: ABCD is a parallelogram ? AE = CF ? aecf is a parallelogram ᙽ AF / / EC ᙽ MF / / en ? AB = DC and AE = CF ᙽ EB = DF ∵ AB / / CD ? ebfd is a parallelogram ? WB / / DF ? EM / / NF ∵ MF / / EN ? EMFN is a parallelogram

In a circle, AB is parallel to CD, CE is the diameter of circle O, BOD = 36 ° and the degree of ACE is calculated

∵AB∥CD
/ / arc AC = arc BD
∵∠BOD=36°
∴∠AOC=36°
∵OA=OC
∴∠ACO=∠CAO=72°
That is, ACE = 72 degrees

It is known that AB, CE are the diameter of circle O, CD is the chord of circle O, CD / / AB, find AC = BD = be

Because the points a, B, C, D, e are all on the circumference, OA = ob = OC = od = OE,
OAC, OBD, OBE and OCD are isosceles
In △ OAC and △ OBD
Because CD / / AB
So ∠ AOC = ∠ DCO
∠BOD=∠CDO
All OCDs are isosceles
So OCD = ∠ ODC
From the above three equations, we can get: ∠ AOC = ∠ BOD
OA = ob OC = OD
So △ OAC and △ OBD are congruent
AC=BD
In △ OAC and △ OBE
OA=OB=OC=OE
∠ AOC = ∠ BOE (opposite vertex angle)
So △ OAC and △ OBE are congruent
AC=BE
Then: AC = BD = be
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