AB is the diameter of circle O, BC is the chord, OD ⊥ BC is at point E, intersection arc BC is at point D. please write five different types of correct conclusions If BC = 8, ed = 2, find the radius of circle o

AB is the diameter of circle O, BC is the chord, OD ⊥ BC is at point E, intersection arc BC is at point D. please write five different types of correct conclusions If BC = 8, ed = 2, find the radius of circle o

The angle c = 90 degrees od / / AC BD arc = DC arc the triangle OBE is similar to the triangle ABC ed * (EO + R) = be * EC
2(2R-2)=4*4 R=5

As shown in the figure, in circle O, CD is the diameter, AB is the chord, ab ⊥ CD is in M. if CD = 10cm, OM: OC = 3:5, find the length of chord ab

Let OC = 5x om = 3x OC = od = 5x OC + od = 5x + 5x = 10x = 10cm x = 1cm OC = od = 5cm om = 3x = 3cm because AB is perpendicular to m, so am is perpendicular to CD to m, so OAM is a right triangle, so am square + OM square = Ao square, so am square = 25-9 = 16cm, so am = 4cm because AB is chord, CD is diameter, ab = 2am = 8cm

It is known that in ⊙ o, CD is the diameter, AB is the chord, ab ⊥ CD is in M, CD = 15cm, if om: OC = 3:5, then ab=______ cm．

As shown in the figure, the connection OA, ∵ CD is the diameter, AB is the chord, ab ⊥ CD is in M, ᚉ am = BM,

As shown in the figure, if the diameter of ⊙ o is CD = 10cm, AB is the chord of ⊙ o, ab ⊥ CD, the perpendicular foot is m, OM: OC = 3:5, then AB is=______ cm．

∵ circle O diameter CD = 10 cm,
The radius of circle O is 5cm, that is, OC = 5cm,
∵OM：OC=3：5，
∴OM=3
5OC=3cm，
Connect OA,
∵AB⊥CD，
/ / M is the midpoint of AB, i.e. am = BM = 1
2AB，
In RT △ AOM, OA = 5cm, OM = 3cm,
According to Pythagorean theorem: am=
OA2−OM2=4cm，
Then AB = 2am = 8cm
So the answer is: 8

In 0 o, CD is the diameter, AB is the chord, AB is vertical CD, CD is 15, OM is 3 to 5, and the length of chord AB is obtained

I think m is the intersection of AB and CD `Because CD = 15 and CD is diameter, OC = 15 / 2
Because om: OC = 3:5, so om = 9 / 2, because AB is perpendicular to CD, according to the vertical diameter theorem, we can get the chord AB = Calculate by yourself

If the diameter of circle O is CD = 10cm, take point m on radius OD, chord ab ⊥ od is at point m, OM; OC = 3:5, then the length of chord AB is

Connect OA and ob
∵CD＝10
∴OC＝OD＝CD/2＝5
∴OA＝OC＝5
∵OM：OC＝3：5
∴OM＝3
∵AB⊥OD
∴AM＝BN＝AB/2
∵AB⊥OD
∴AM²＝OA²-OM²＝25-9＝16
∴AM＝4
∴AB＝2AM＝8

If the diameter ab of O intersects with the chord CD at point E, given AE = 6cm, EB = 2cm, ∠ CEA = 30 °, then the length of chord CD is () A. 8cm B. 4cm C. 2 Fifteen D. 2 Seventeen

O is used as om ⊥ CD, and OC is connected,
∵AE=6cm，EB=2cm，
∴AB=8cm，
∴OC=OB=4cm，
∴OE=4-2=2（cm），
∵∠CEA=30°，
∴OM=1
2OE=1
2×2=1（cm），
∴CM=
OC2−OM2=
42−12=
15，
∴CD=2CM=2
15．
Therefore, C

As shown in the figure, in ⊙ o, AB is the diameter, chord CD ⊥ AB, e is a point on arc BC. If ⊙ CEA = 28 °, then ﹤ bad=______ .

∵ AB is the diameter,
∴∠ADB=90°，
∵ chord CD ⊥ AB,
Qi
AC=
AD，
∴∠CEA=∠B=28°，
∴∠BAD=90°-∠B=62°．
So the answer is: 62 degrees

It is known that AB is the diameter of ⊙ o, and the chords CD ⊥ AB, e are At a point on AC, the extension lines of AE and DC intersect at point F. it is proved that: ∠ AED = ∠ CEF

Proof: connect ad, as shown in the figure,
∵CD⊥AB，
/ / arc AC = arc ad,
∴∠ADC=∠AED，
∵∠CEF=∠ADC，
∴∠AED=∠CEF．

It is known that the chords AB and CD of ⊙ o intersect at point E, The degree of AC is 60 degrees, If the degree of BD is 100 ° then ∠ AEC is equal to______ .

Connect BD and BC, as shown in the figure,
A kind of
The degree of AC is 60 degrees,
The degree of BD is 100 degrees,
∴∠ABC=1
2×60°=30°，∠BCD=1
2×100°=50°，
∵∠AEC=∠EBC+∠ECB，
∴∠AEC=30°+50°=80°．
So the answer is 80 degrees