AB is the diameter of circle O, the chord AC is parallel to the radius OD, proving that CD = dB

AB is the diameter of circle O, the chord AC is parallel to the radius OD, proving that CD = dB

Connect OC
AC ‖ OD, so ∠ OCA = ∠ cod, ∠ OAC = ∠ BOD
Because of the OAC = OCA, the COD = BOD can be obtained
Therefore, OB = OC, OD = OD, and △ OCD is equal to △ OBD, so CD = dB

In the circle O, the chords AB and CD are perpendicular to each other, intersecting the circle O at four points a, B, C and D, connecting OA, ob, OC and OD. It is proved that the angle AOD + angle BOC = 180 degrees

The corresponding relationship between the inner angle of a circle and the corresponding arc is used to prove that:
1) The strings AB and CD are perpendicular to each other,
==(AD arc + CB arc) / 2 = 90 degrees
==(AD arc + CB arc) = 180 degrees
2) Angle AOD + angle BOC = (AD arc + CB arc) = 180 degrees

In ⊙ o, the length of diameter AB is 6, OD ⊥ chord AC, D are perpendicular feet, BD and OC intersect at point E, then the length of OE is______ .

Connect BC,
According to the meaning of the title, draw the picture
∵ AB is the diameter,
∴∠ACB=90°，
∵ OD ⊥ chord AC, D is perpendicular foot,
∴DO∥BC，
∴AD=CD，DO=1
2BC, (the median line theorem of triangles)
∴△DOE∽△BCE，
∴DO
BC＝EO
EC=1
2，
∵AB=6，
∴CO=3，
The length of OE is 1
So the answer is: 1

As shown in the figure, AB is a chord of circle O, OD ⊥ AB, perpendicular foot is C, intersect circle O at point D, OC = 3, OA = 5, find the length of ab

Because the angle OCA = the angle OCB = the right angle, according to the Pythagorean theorem, the square of Ao = the square of CO + the square of AC, so AC is equal to 4, so AB = 2Ac = 8

In the circle O, the diameter AB is 6, the OD vertical chord AC is at D, BD and OC are at e, and OE is obtained

The diameter of the circle is the diameter of the circle, and AC is the chord ᙨ BC ? AC ∵ ACB = 90 ? OD ⊙ AC ? AC ? ADO = 90  OD ? BC ? OED ∵ and ∵ BEC is the vertex angle ? the ᙨ OED ? CEB is the similar triangle ? OE / CE = od / BC M OA = OC ? OD 88; AC Δcod is the phase

In circle O, the length of diameter AB is 12cm, OD is the chord center distance of chord AC, BD and OC intersect at point E, then the length of CE is____ Cm.

Connect BC
∵ diameter ab
∴∠ACB＝90
∵OD⊥AC
∴∠ADO＝90,AD＝CD
∴∠ADO＝∠ACB
∴OD∥BC
The OD is the median line of △ ACB
∴OD/BC＝1/2
And ∵ OD ∫ BC
∴OE/CE＝OD/BC＝1/2
∵OC＝AB/2＝12/2＝6
∴OE＝OC-CE＝6-CE
∴（6-CE）/CE＝1/2
∴CE＝4（cm）

As shown in the figure, ab = CD, AE ⊥ BC in E, DF ⊥ BC in F, CE = BF, connect ad, intersect EF at point O, guess which line segment o is the midpoint? Please choose one of the conclusion proof

In △ AEB and △ DFC, be = CF, ∠ AEB = ∠ CFD = 90 °, ab = CD, ∧ AEB ≌ △ CFD (SAS),  AE = DF. ∵ AE ⊥ BC, ⊥ BC, ⊥ AE ⊥ BC, ⊥ AE ⊥ DF, ≌≌≌△ CFD (SAS),  AE = DF.

As shown in the figure, ab = CD, AE ⊥ BC in E, DF ⊥ BC in F, ad crossing EF in O, OA = OD, verification: BF = CE

∵ Ao = OD,  AOE = ∠ DOF (opposite vertex angle), ∠ AEO = ∠ DFO = 90 °
∴△AEO≌△DFO
∴DF=AE,OF=OE
In RT △ AEB and RT △ DFC
AB=CD,AE=DF
∴△AEB≌△DFC（HL）
∴BE=CF
∴BE-EF=CF-EF
CE = BF

As shown in the figure, ab = CD, AE ⊥ BC in E, DF ⊥ BC in F, CE = BF, connect ad, intersect EF at point O, guess which line segment o is the midpoint? Please choose one of the conclusion proof

In △ AEB and △ DFC, be = CF, ∠ AEB = ∠ CFD = 90 °, ab = CD, ∧ AEB ≌ △ CFD (SAS),  AE = DF. ∵ AE ⊥ BC, ⊥ BC, ⊥ AE ⊥ BC, ⊥ AE ⊥ DF, ≌≌≌△ CFD (SAS),  AE = DF.

It is known that AB is parallel to CD, ad crosses BC at point O, EF passes through point O, respectively crosses AB CD at point E F and AE = DF, it is proved that O is the midpoint of EF (the process is complete, the reason will be written later

Because AB is parallel to CD
The angle ead = angle ADF, angle AEF = angle DFE, (the two lines are parallel, and the internal staggered angle is equal)
AE = DF,
According to the angle, side angle (two corners and their clamped edges correspond to two equal triangles congruent)
The AoE of triangle is equal to DOF of triangle
So EO = of
So o is the midpoint of EF
If satisfied, please accept, thank you!