As shown in the figure, AB and CD are the two chords of circle O. extending Ba and DC respectively, intersecting at points P, m, n are the midpoint of arc AB and arc CD, and Mn ⊥ Po Speed, please

As shown in the figure, AB and CD are the two chords of circle O. extending Ba and DC respectively, intersecting at points P, m, n are the midpoint of arc AB and arc CD, and Mn ⊥ Po Speed, please

Proof: even OM, ON.OM Intersection AB at point R, on intersect CD at point t,
Because OM, on is the radius of the circle
So om = on,
Because Po ⊥ Mn
So ∠ mop = ∠ NOP (three wire in one)
Because m, n are the midpoint of arc AB and arc CD
So om ⊥ AB, on ⊥ CD
So ∠ BPO = DPO
So or = ot (the distance from the point on the bisector to both sides of the corner is equal)
So AB = CD (in the same circle, the center of the circle is equal to the chord opposite)

Known: as shown in the figure, M is Let the radius of ⊙ o be 4cm and Mn = 4 3cm． (1) Find the distance from the center O to the chord Mn; (2) Find the degree of ∠ ACM

(1) Connect to OM,
∵ point m is
AB of the midpoint,
∴OM⊥AB，
Through point O, make OD ⊥ Mn at point D,
According to the vertical diameter theorem, MD = 1
2MN=2
3，
In RT △ ODM, OM = 4, MD = 2
3，
∴OD=
OM2−MD2=2，
Therefore, the distance from center O to chord Mn is 2cm;
（2）cos∠OMD=MD
OM＝
Three
2，
∴∠OMD=30°，
∵ m is the midpoint of the arc AB, and OM passes through o,
∴AB⊥OM，
∴∠MPC=90°，
∴∠ACM=60°．

Given that the diameter of circle O is 4cm, C is the midpoint of arc AB, the chord AB / CD intersects point P, CD = 2 √ 3cm, calculate the degree of angle APC

According to the number relation in the question, the angle OCD = angle OCP = 30 ° because C is the midpoint of arc AB, so OC is perpendicular to chord AB, so angle APC = 60 °

Given that the diameter of the circle O is 4cm, the chords AB and CD intersect at the point P, C is the midpoint of the arc AB, CD = (2 times the root sign 3) cm, calculate the degree of angle APC

30 degrees

The chord AB, CD of ⊙ o intersects point P, PA = 4, Pb = 3, PC = 6, AE cuts ⊙ o at point a, the extension line of AE and CD intersects point E, EA = 2, root sign 5, and find the length of PE

According to secant theorem, PA * Pb = PC * PD, so PD = PA * Pb / PC = 4 * 3 / 6 = 2
According to the cutting line theorem: AE ^ 2 = ed * EC, therefore, 20 = ed * (de + PD + PC) = ed (ED + 8),
The solution is: ed = 2, (ED = - 10, omitted)
Therefore, PE=ED+PD=2+2=4

As shown in the figure, in ⊙ o, AB is the diameter, CD is the chord, ab ⊥ CD (1) P is What is the relationship between ∠ cpd and ∠ cob on CAD (not coincident with C and D)? Try to explain the reasons; (2) The point P 'is at What is the quantitative relationship between ∠ CP ′ D and ∠ cob on CD (not coincident with C, d)? Why?

（1）∠CPD=∠COB．… (1 point)
Reason: connect od as shown in the figure (2 points)
∵ AB is the diameter, ab ⊥ CD,
Qi
BC=
BD，… (3 points)
∴∠COB=∠DOB=1
2∠COD．… (4 points)
And ∵ cpd = 1
2∠COD，
∴∠CPD=∠COB… (5 points)
(2) The quantitative relationship between ∠ CP'd and ∠ cob is ∠ CP'd + ∠ cob = 180 ° (6 points)
Reason: ∵ cpd = 1
2∠COD，∠CP'D=1
2（360°-∠COD）=180°-1
2∠COD，
∴∠CPD+∠CP'D=180°．… (8 points)
It is known from (1) that ∠ cpd = ∠ cob,
∴∠CP'D+∠COB=180°．… (9 points)

In the same circle, chord AB = chord CD, compare the length of arc AB and arc CD, and prove the conclusion I know the result of this problem, but I can't write the proof process

Connect OA ob OC OD
OA=OC OB=OD AB=CD
Triangle OAB is equal to triangle OCD
Angle AOB = angle cod
Then the length of arc AB and arc CD are equal

In the circle O, if two chords AB, CD, AOC = 30 ° and BOD = 70 ° and the intersection point of AB and CD is e, then the angle AEC =?, Double problem solving, main picture and problem solving process, fast

Firstly, a theorem is proved
"The degree of the angle of the vertex in the circle (both sides intersect the circle) is equal to half of the sum of the degrees of the two arcs it cuts"
prove:
Pass C as CP / / AB, intersect with P,
Then ∠ AEC = ∠ C, arc AC = arc BP (the arcs between two parallel chords in the circle are equal)
The degree of ∠ C is equal to half of arc DP, and arc DP = arc BD + arc BP = arc BD + arc AC
So the degree of AEC is equal to half of "arc BD + arc AC"
That is, "the degree of the angle of the vertex in the circle (both sides intersect the circle) is equal to half of the sum of the degrees of the two arcs it cuts."
Ψ AEC = 1 / 2 (arc AC + arc BD) = 1 / 2 (∠ AOC + ∠ BOD) = 1 / 2 (30 ° + 70 °) = 50 °

As shown in the figure, in the concentric circle with o as the center, the chord ab of the large circle intersects the small circle at C and D （1）∠AOC=∠BOD；（2）AC=BD．

(1) It is proved that O is OE ⊥ ab,
∵ OAB and △ OCD are isosceles triangles,
∴∠AOE=∠BOE，∠COE=∠DOE，
∴∠AOE-∠COE=∠BOE-∠DOE，∠AOC-∠BOD；
(2) Proof: ∵ OE ⊥ AB,
∴AE=BE，CE=DE，
ν be-de = ae-ce, i.e. AC = BD

As shown in the figure ···, Ao is perpendicular to Bo, CO is perpendicular to do, angle AOC is equal to 1:5, then angle BOD is equal to? Five is necessary,

When OC is between AB, ∠ AOC = 1 / 5 * ∠ AOB = 18 °, BOD = 18 ° or 162 °
When OC and ab are outside, AOC = 1 / 4 * ∠ AOB = 22.5 ° and BOD = 22.5 ° or 157.5 ° respectively