As shown in Figure AB is the diameter of circle O, ad and circle O are tangent to point a, de and circle O are tangent to point E, and point C is a point on the extension line of De, and CE = CB is connected to the extension of AE and AE

: (1) connect OE, OC; (1 point)
∵CB=CE,OB=OE,OC=OC
∴△OEC≌△OEC（SSS）
﹤ OBC = ∠ OEC (2 points)
And ∵ de and ⊙ o are tangent to point E
﹤ OEC = 90 ° (3 points)
∴∠OBC=90°
⊙ BC is the tangent line of ⊙ O. (4 points)
(2) Pass point D as DF ⊥ BC at point F,
∵ ad, DC, BG cut ⊙ o at point a, e, B respectively
∴DA=DE,CE=CB,
Let BC be x, then CF = X-2, DC = x + 2,
In RT △ DFC,,
(6 points)
∵AD‖BG,
∴∠DAE=∠EGC,
∵DA=DE,
∴∠DAE=∠AED；
∵∠AED=∠CEG,
∴∠EGC=∠CEG,
/ / CG = CE = CB =, (7 points)
∴BG=5,
ν Ag =; (8 points)
Solution 1: connect be,,
∴ ,
(9 points)
In RT △ Berg,
, (10 points)
Solution 2: ? DAE = ∠ EGC,  AED = ∠ CEG,
﹤ ade ∽ GCE, (9 points)
(10 points)

It is known that ab = ad, CB = CD, e and F are the midpoint of AB and ad respectively

Proof: connect AC,
In △ ABC and △ ADC,
AB＝AD
CB＝CD
AC＝AC ，
∴△ABC≌△ADC（SSS），
∴∠B=∠D，
E and F are the midpoint of AB and ad respectively,
∴BE=1
2AB，FD=1
2AD，
∵AB=AD，
∴BE=FD，
In △ BEC and △ DFC,
BE＝FD
∠B＝∠D
BC＝DC ，
∴△BEC≌△DFC（SAS），
∴CE=CF．

As shown in the figure: AC= CB, D and E are the midpoint of radius OA and ob respectively, Confirmation: CD = CE

Proof: connect OC
In ⊙ o, ⊙ o, ∵
AC=
CB
∴∠AOC=∠BOC，
∵ OA = ob, D and E are the midpoint of radius OA and ob respectively,
∴OD=OE，
∵ OC = OC (common side),
∴△COD≌△COE（SAS），
The corresponding sides of an congruent triangle are equal

As shown in the figure, D and E are the midpoint of ⊙ o radius OA and ob respectively, and C is the Is CD equal to CE at the midpoint of AB? Why?

CD = CE for the following reasons: (1 point)
Connect OC,
∵ D and E are the midpoint of ⊙ o radius OA and ob respectively,
∴OD=1
2AO，OE＝1
2BO，
∵ OA = ob, ᙽ od = OE, (2 points)
∵ C is
The midpoint of AB
AC＝
BC，
Ψ AOC = ∠ BOC, (4 points)
≌△ Eco, (5 points)
ν CD = CE. (6 points)
So the answer is: CD = CE

As shown in the figure, D and E are the radius of circle O OA, points on ob respectively, CD ⊥ OA, CE ⊥ ob, CD = CE

prove:
Connect OC
∵CD⊥OA,CE⊥OB
∴∠CEO=∠CDO=90º
And ∵ CD = CE, OC = OC
∴Rt⊿CEO≌Rt⊿CDO(HL)
∴∠AOC=∠COB
The arc AC = arc CB [the arc corresponding to the center angle of an equal circle in the same circle is equal]

As shown in the figure: AC= CB, D and E are the midpoint of radius OA and ob respectively, Confirmation: CD = CE

In ⊙ o, arc AC = arc CB, D and E are the midpoint of radius OA and ob respectively. It is proved that CD = CE is fast

prove:
Connect Ca, CB, make CF perpendicular to AB, cross AB to F, and cross De to g
Because D and E are the midpoint of radius OA and ob respectively
So De is parallel to ab
Because CF is perpendicular to ab
So CF is perpendicular to de
Because OA = ob
So CF is the bisector of the base of the isosceles triangle ode,
So CF is the vertical bisector of de
So CD = CE

It is known that CD is the height on the edge of triangle ab. circle O with diameter CD intersects CA respectively. CB and point G are the midpoint of AD. it is proved that CE is tangent of circle o

It is known that CD is the height on the edge of triangle AB, and the circle O with CD as its diameter intersects Ca, CB at e and f respectively, and point G is the midpoint of AD
Let the midpoint of CD (i.e. the center of circle O) be h, and connect he and de,
Then ∠ Dec = ∠ DEA = 90 °
In RT △ ade,
∵ G is the midpoint of AD,
/ / EG = DG = Ag (center line on the hypotenuse of a right triangle)
∵ eh, DH are the radius of circle o,
∴EH=DH,
And GH is the public side,
∴△EGH≌△DGH
But CD ⊥ AB,
∴∠GEH=∠GDH=90°,
∴GE⊥EH
﹤ Ge is the tangent line of circle O (the straight line passing through a point on the circle perpendicular to the radius passing through this point is the tangent line of the circle)

As shown in the figure, in the quadrilateral ABCD, AC bisects the angle bad, makes CE and AB in e through C, and CD = CB, angle ABC + angle ADC = 180 ° 0, and AE = 1 / 2 (AB + AD)

It is proved that the extension of CF ⊥ ad to F
If AC bisection ∠ bad; CE ⊥ AB, then CF = CE
And CD = CE, so RT ⊿ CFD ≌ RT Δ CEB (HL), DF = be;
If AC = AC, then RT ⊿ AEC ≌ RT Δ AFC (HL), then AE = AF
So: AE = (AE + AF) / 2 = [(ab-be) + (AD + DF] / 2 = (AB + AD) / 2
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As shown in the figure, ad and CE are higher than △ ABC, and ab = 2BC? Why?

AD=2CE．
The reasons are as follows: s △ ABC = 1
2AB•CE=1
2BC•AD，
∵AB=2BC，
∴1
2•2BC•CE=1
2BC•AD，
Ad = 2ce

As shown in Figure AB is the diameter of circle O, ad and circle O are tangent to point a, de and circle O are tangent to point E, and point C is a point on the extension line of De, and CE = CB is connected to the extension of AE and AE