1. When connecting AC BD and CO DO, AC=CO DO=DB is because CM is vertically divided by AO DN and BO, and AC= CO=DO = DB is because CO=DO (radius), that is, AC=DB, so the arc AC = BD.

2. Let the midpoint of AB be m, connect OM, and extend the intersection of OM to n..

Then am = BM = CD

Make an arc with A and B as the center and AM and BM as the radius.

Then there must be intersections e ANd f on arcs an and BN, so arc AE = arc BF = arc CD because chord AE = BF = ab/2 = CD.

Because arc AB = arc AE+ arc EF+ arc BF = 2 arc CD+ arc EF, arc AB > 2 arc CD.

(it cAN also be proved by connecting an: because an > am = CD, in △AON and △COD, because OA = of = OC = OD, and an > CD.

So ∠ AON > ∠ COD knows arc An > arc CD according to the arc length formula, so arc AB > 2 arc CD).

3. Let ∠ AOD = X ∠ ODE = X, ∠ OCD = 180-2x, ∠ OCE = 80 degrees, ∠ OED = ∠ ODE = X, ∠ DOE = 65436.

You can get 3∠AOD=∠BOE.

4.( 1) Try to explain AE=BF.

Make OG perpendicular to CD, CD cross with G, and connect OC with OD and DE to get CG=GD. OE=OF is obtained by the theorem that parallel lines are cut into proportional line segments, and AE=BF is obtained by simple substitution.

(2) During the sliding process of the moving string CD, the area of the quadrilateral CDFE remains unchanged.

Always equal to 2 * (EC+DF) * CD * 2 = (EC+DF) * CD = og * CD.

Multiple choice question:

1.A

2.A