In the circle O, the radius is 4 and the angle AOB is 60 degrees. Point C is the midpoint of arc AB, CM is vertical OA, CN is vertical ob, and the perpendicular feet are points m and N, respectively (1) (2) when point C moves on arc AB, try to judge whether the length of Mn changes and prove your conclusion (1) (2) when point C moves on arc AB, try to judge whether the length of Mn changes and prove your conclusion

In the circle O, the radius is 4 and the angle AOB is 60 degrees. Point C is the midpoint of arc AB, CM is vertical OA, CN is vertical ob, and the perpendicular feet are points m and N, respectively (1) (2) when point C moves on arc AB, try to judge whether the length of Mn changes and prove your conclusion (1) (2) when point C moves on arc AB, try to judge whether the length of Mn changes and prove your conclusion

(1) Since C is the midpoint of arc AB, then ∠ MOC = ∠ cob (the center angles of equal arcs are equal)
If cm is perpendicular to OA and cn is perpendicular to ob, then △ OMC ≌ △ onc can be known
Then om = on and ∠ AOB = 60 degrees, then △ omn is an equilateral triangle
If OC = 4, AOC = 30, then Mo = 2 (√ 3) = Mn
(2) The length of Mn remains unchanged. Extend ob, AC to h, make ∠ bog = 90 degree to cm to g, make MF ⊥ go to F, me ⊥ OB to E
Because ∠ AOB = 60 degrees, ∠ CMO = 90 degrees, ∠ CNO = 90 degrees, ∠ gob = 90 degrees,
Then cn ∥ go, ∠ OGM = 30 degrees, ∠ g = 60 degrees, ∠ H = 30 degrees. Let on = x, CN = √ (16-x ^ 2) be known from CO = 4
NH = (√ 3) × CN = √ (48-3x ^ 2) = y
Then Oh = x + y, then MH = (√ 3 / 2) × Oh = (√ 3) (x + y) / 2
Then me = Oh / 2 = (√ 3) (x + y) / 4, OE = (√ 3 / 3) × me = (x + y) / 4
Then en = on-oe = x - (x + y) / 4 = (3x-y) / 4
EN^2+ME^2=[(3x-√(48-3x^2))^2]/16+[(√3)(x+√(48-3x^2))/4]^2
=Add [9x ^ 2-6x √ (48-3x ^ 2) + 48-3x ^ 2] / 16 + [3x ^ 2 + 6x √ (48-3x ^ 2) + 3 × 48-9x ^ 2] / 16 molecules, and combine the similar terms
=[(9-3+3-9)x^2+(6x-6x)√(48-3x^2)+48+144]/16
=[144+48]/16=12=MN^2
Then Mn is the fixed value, which is √ (12) = 2 (√ 3)
(although it's complicated, it's the best I can do now. I hope I can understand it.)

As shown in the figure, D and E are respectively the midpoint of the radius OA and ob of ⊙ o, and point C is the midpoint of ab

It is proved that: ∵ point C is the middle point of AB, ∵ AOC = ∵ BOC; ∵ D and E are the middle points of radius OA and ob of ⊙ o, ∵ od = OE = 12oa; and ∵ OC = OC, ≌ cod ≌ COE (SAS);; CD = CE

As shown in the figure, D and E are respectively the midpoint of the radius OA and ob of ⊙ o, and point C is the midpoint of ab

It is proved that: ∵ point C is the middle point of AB, ∵ AOC = ∵ BOC; ∵ D and E are the middle points of radius OA and ob of ⊙ o, ∵ od = OE = 12oa; and ∵ OC = OC, ≌ cod ≌ COE (SAS);; CD = CE