It is known that: as shown in the figure, take the oblique edge ab of RT △ ABC as the diameter to make ⊙ o, and D is the point on ⊙ o, and there is AC = CD. Pass through point C to make the tangent of ⊙ o, and connect CD with the extension line of BD at point E (1) Try to judge whether be and CE are perpendicular to each other, please explain the reason; (2) If CD = 2 5，tan∠DCE=1 2. Find the radius length of ⊙ o

(1) ∵ AB is the diameter
∴∠ACB=90°
∵AC=CD，
∴∠ABC=∠CBE，
∵ CE is the tangent of ⊙ o,
∴∠BCE=∠A，
∴∠BEC=∠ACB=90°
∴BE⊥CE．
(2) ∵ CE is tangent, AC = CD,
∴∠DCE=∠DBC=∠ABC，tan∠DCE=1
Two
∴tan∠ABC=1
Two
∵AC=CD=2
Five
∴BC=4
Five
∴AB=10
The radius of ⊙ o is equal to 5

In ⊙ o, AC= CE． (1) As shown in Figure 1, Co ⊥ AE was confirmed; (2) As shown in Fig. 2, the diameter of CD ⊥ AB is D, if BD = 1, AE = 4, find the radius of ⊙ o

(1) It is proved that: extending co intersection AE at point D, ∵ AC = CE, CD passing through the center, ∵ Co ⊥ AE; (2) let the radius of ⊙ o be r, connect CO and extend the intersection of CO with point F, ∵ AC = CE, CF over the center, AE = 4,  of ⊥ AE, ? AF = 12 × 4 = 2, ∵ CD ⊥ AB, ∵ AOF = ≁ cod, ? in ⊥ oaf and ⊥ OCD, ?

As shown in the figure, AB is the diameter of circle O, inferior arc BC arc = be arc, BD / / CE, connect AE and extend the intersection BD at point D. verify that the square of AB = AC times ad

It is proved that: (1) ∵ inferior arc BC arc = be arc,
∴∠1=∠2,
Inferior arc AC = inferior arc AE, AC = AE
∴AB⊥CE．
∵CE∥BD,∴AB⊥BD．
⊙ BD is the tangent line of ⊙ o
(2) Connect CB
∵ AB is the diameter of ⊙ o, ᙽ ACB = 90 °
∵∠ABD=90°,∴∠ACB=∠ABD．
∵∠1=∠2,∴△ACB∽△ABD．
∴AC/AB=AB/AD,AB²•AC

Known: as shown in the figure, AB is the diameter of ⊙ o, points c and D are on ⊙ o, CE ⊥ AB is on e, DF ⊥ AB is on F, and AE = BF, is AC equal to BD? Why?

AC is equal to BD
The reasons are as follows:
Connect OC and OD, as shown in the figure,
∵OA=OB，AE=BF，
∴OE=OF，
∵CE⊥AB，DF⊥AB，
∴∠OEC=∠OFD=90°，
In RT △ OEC and RT △ ofd,
OE＝OF
OC＝OD ，
∴Rt△OEC≌Rt△OFD（HL），
∴∠COE=∠DOF，
/ / AC arc = BD arc,
∴AC=BD．

As shown in the figure, A.B.C is three points of circle O, arc AC = arc BC, point m is a point on BC, CE ⊥ am, AE = 5, M = 3, find BM

Connect cm, intercept AF = BM on AE, and connect cm
∵∠A=∠B,AF=BM,CA=CB
∴△ACF≌△BCM
∴CF=CM
∵CE⊥FM
∴EF=EM
∴AF+EF=EM+BM
AE = EM + BM
∵AE=5,EM=3
∴BM=2

As shown in the figure, AB is the diameter of circle O, point C is the midpoint of arc AB, chord CE intersects AB at point F, D is a point on the extension line of ab, If of = 1, OA = 3, calculate the area of △ ace

connect OC.AB Let BD = x, then de = DF = 2 + X. De is the tangent line of the circle, then: de = BD * ad, (2 + x) mm2 = x (x + 6), x = 2. That is, BD = 2, de = 4. Connect OE, OE ⊥ De to make eh ⊥ od in H. from the area relationship, we can know that de * OE = od * eh

It is known that AB and CD are the two diameters of a circle. The chord CE is parallel to ab. the degree of arc CE is 40 degrees. Find the degree of ∠ BOD. (proof process) fast!

∵ the degree of the arc CE is 40 degrees
That is, COE = 40
AB / / CE
∴∠AOC=∠COE=40
∵ AOC and  BOD are antiparietal angles
∴∠BOD=∠AOC=40
Floor length, I'm a word by word, give bonus points^-^

As shown in the figure, AB and CD are the diameters of circle O, the chord CE is parallel to AB, ∠ BOD = 110 ° and the degree of arc CE is calculated?

If OE is connected, OE = OC, so ∠ OEC = ∠ OCE
Because ∠ BOD = 110 °, so ∠ BOC = 70 °
CE parallel ab
∠OCE＝∠BOC＝70°,∠OEC＝70°
In triangle Oce
∠COE=180°－（∠OCE+∠OEC）＝40°
That is, the degree of arc CE is 40 degrees

As shown in the figure, the extension lines of the two chords Ba and CE of circle O intersect a point D outside the circle, and the connecting do intersects the Circle F, and the arc EF = arc FA. It is proved that EC = ab

Proof: connect OA and OE, make op ⊥ AB to P, OQ ⊥ CE to Q
Arc EF = arc AF, so ∠ AOF = ∠ EOF
In △ AOD and △ EOD
∠AOF=∠EOF
AO=EO
OD=OD
So △ AOD ≌ △ EOD. ∠ ADO = ∠ Edo
OP⊥AB,∠OPD=90
OQ⊥CE,∠OQD=90
In RT △ OPD and RT △ OQD,
∠ADO=∠EDO
∠OPD=∠OQD=90
OD=OD
So RT △ OPD ≌ RT △ OQD (HL). OP = OQ
In the same circle, the distance from the center O to the two strings AB and CE is equal
So AB = CE

Given that AB and CD are the two diameters of circle O, the chord CE is parallel to AB, and the degree of arc CE is 40 °, then ∠ BOD =?

Should be 70 or 110 degrees! Because a, B two points can be interchanged!