It is known that: in △ AOB and △ cod, OA = ob, OC = OD, ∠ AOB = ∠ cod = 90 ° It is known that: in △ AOB and △ cod, OA = ob, OC = OD, ∠ AOB = ∠ cod = 90 °. (1) as shown in Figure 1, points c and D are respectively on the sides OA and ob, connecting AD and BC, and point m is the midpoint of line BC and connecting OM, then the quantitative relationship between line AD and OM is, and the positional relationship is; (2) as shown in Fig. 2, rotate △ COD in Fig. 1 anticlockwise around point o with rotation angle of α (0 ° < α < 90 °), Point m is the midpoint of line BC, connecting OM. Please judge whether the two conclusions in (1) still hold. If yes, please prove; if not, please explain the reasons; (3) as shown in Figure 3, rotate △ COD in Figure 1 anticlockwise around point O to make the OD of △ cod and the edge OA of △ AOB exactly on the same line, point C falls on ob, Point m is the midpoint of line BC. Please judge whether the quantitative relationship between line AD and OM in (1) has changed, write your conjecture and prove it

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Take any point o on the line AB, and pass through the point O to make OC ⊥ OD. When ∠ AOC = 30 °, the degree of ∠ BOD is 0______ .

When OC and od are on the same side of line AB, as shown in the figure:
∵OC⊥OD，∠AOC=30°；
∴∠BOD=180°-∠COD-∠AOC=180°-90°-30°=60°；
When OC and od are on the opposite side of line AB, as shown in the figure:
∵OC⊥OD，∠AOC=30°；
∴∠BOD=180°-∠AOD=180°-（∠DOC-∠AOC）=180°-（90°-30°）=120°．

The radius of circle O is 12 and the chord AB is 16. When the two ends of chord AB slide on the circumference, what kind of figure does the midpoint of chord AB form

For the same circle, the distance from the center of the circle to all chords of the same chord length is equal, that is, the distance from the midpoint of the chord with the same chord length to the center of the circle is equal, so the midpoint of string AB forms a circle with o as its center
Circle with o as the center and 4 * radical 5 as the radius
If the midpoint of AB is set as m, then the OM length is always 4 times the root 5

Known: diameter ab ⊥ chord, Cd in E, ae-be = 16cm, CD = 12cm, then ab=

Because (CD / 2) 2 = AE × EB, AE × EB = 36, and ae-be = 16, we can get AE = 18cm, be = 2cm
So AB = AE + be = 20cm

If there are two known chord ABCD = 20cm in diameter, ABCD = 16cm

(12+16)*(2+1)
=28*3
=84

The radius of circle O is 5 and the length of chord AB is 8

AC = 6 (extend Ao intersection circle O as point D, ad = 10, ab = 8, according to Pythagorean theorem, BD = 6

If the chord ab ∥ CD, ab = 6cm, CD = 8cm in a circle with a radius of 5cm, then the distance between chord AB and CD is______ .

Pass through point o as OE ⊥ AB at E
∵AB∥CD，∴OF⊥CD
∵ OE over center, OE ⊥ ab
∴EB=1
2AB=3cm
∵OB=5cm，∴EO=4cm
Similarly, of = 3cm
∴EF=1cm
When AB and CD are located on both sides of the circle center, EF = 7cm
ν EF = 1cm or EF = 7cm

As shown in the figure, AB is the diameter of ⊙ o, and CD is a chord. If AB = 10cm and CD = 8cm, then the sum of the distances between a and B and the straight line CD is______ .

In RT △ OCG, OC = 12ab = 5cm, CG = 4cm. According to the Pythagorean theorem, we get that: og = oc2 − CG2 = 3cm, AE ⊥ EF, og ⊥ EF, BF ⊥ EF, ᙽ AE ⊥ og ⊥ BF, and O is the middle point of ab

EF is the diameter of the circle, AB and CD are the two chords in the circle, and ab = 6cm, CD = 8cm, EF = 10cm, CD | AB, connect EB and CF, and calculate the arc area of EAB and CDF At the same time, AB and CD are on both sides of EF, and AB / / CD / / EF

It's simple,
The distance between AB and EF can be calculated by Pythagorean theorem, which is 4,
So the area of EAB is 1 / 2 * 6 * 4,
The distance between Cd and EF can also be calculated by Pythagorean theorem, which is 3,
So the area of CDF is 1 / 2 * 8 * 3,
The sum of area is 24

As shown in the figure, the diameter of ⊙ o is 10cm, the chord AB = 8cm, and P is a moving point on string ab (1) The diameter of ⊙ o is 10cm, chord AB = 8cm, P is a moving point on string AB, and the length range of OP is obtained (2) AB is the chord of ⊙ o, the radius OC, OD intersect AB at point E, F, and AE respectively= BF.OE What does it have to do with the size of of of? Why? ... but in my documents

First answer your first question: the range of OP is 3cm ~ 5cm, 3cm is that point P is located at the intersection of vertical bisector of circle and ab. as for how to get this value, it is to find the length of OP in right triangle OPA or OPB. Pythagorean theorem is easy to get. 5cm means that point P coincides with point a or B, 5cm is the length of radius