It is proved that: 1, AE = BD 2, if AC is vertical be, prove AD + BD = CD × Radix 2

It is proved that: 1, AE = BD 2, if AC is vertical be, prove AD + BD = CD × Radix 2

1. Prove congruence by corner and edge, then come out
2. You may have miscopied the title, it should be AC vertical BC
According to 1, AD + BD = De, CE = CD, the equation comes out
AC vertical be is not true

It is known that in the circle O, the chord AB and the chord CD intersect at point P, and arc AC = arc BD

It happens that I also do this problem
OE ⊥ AB in E, of ⊥ CD in F
∵ arc AC = arc BD, arc ad = arc ad
ν arc CD = arc AB
∴CD=AB
∴OE=OF
Using h. l to prove RT △ OEP ≌ RT △ of P
Po bisection ∠ CPB is obtained

Known: as shown in the figure, in ⊙ o, the chord AB = CD It is proved that: (1) arc AC = arc BD; （2）∠AOC=∠BOD．

It is proved that: (1) in ⊙ o, the chord AB = CD,
/ / arc AB = arc CD,
∵ arc BC = arc CB,
ν arc AC = arc BD;
(2) ∵ arc AC = arc BD,
∴∠AOC=∠BOD．

AB, CD are the chords of circle O and ab ‖ CD. It is proved that arc AC = arc BD AB and CD are ipsilateral and ab < CD

Connect ad, BC
∵AB‖CD
∴∠ABC=∠BCD
∴∠ABC=∠ADC
∴∠BCD=∠ABC
/ / arc AC = arc BD

In circle O, chord AB intersects chord CD at point E, and arc AC = arc BD

prove:
∵ arc AC = arc BD
/ / arc AC + arc BC = arc BD + arc BC
That is, arc AB = arc CD
ν AB = CD [equal arc and equal chord]
The distances from point O to AB and CD are equal [equal chord, equal chord center distance]

In circle O, string AB is parallel to string CD. How to prove that arc AC = arc BD?

It is proved that if we connect ad angle ADC = angle bad (AB parallel CD, equal staggered angle), then arc AC = arc BD (in the same circle, the circumference angle is equal, the opposite arc is also equal)

AB is the diameter of circle O, BC is the chord, OD is perpendicular to BC to e, intersect circle to D, and connect OC. If BC = 8, ed = 2, find the radius of circle o

Let the radius of a circle be x, then OE = od-ed = X-2; OC = x; CE = BC / 2 = 4 triangle, OEC is a right triangle. X = 5 can be solved by Pythagorean theorem (X-2) 2 + 42 = X2, so the radius of circle O is 5

As shown in the figure, AB is the diameter of ⊙ o, BC is the chord, OD ⊥ BC is at e, BC is at d.bc = 8, ed = 2, then the radius of ⊙ o is______ .

Let the radius of ⊙ o be r,
∵OD⊥BC，
∴CE=BE=1
2BC=1
2×8=4，
In RT △ BOE, OE = od-de = R-2, OB = R, be = 4,
∵OE2+BE2=OB2，
∴（R-2）2+42=R2，
The solution is r = 5,
The radius of ⊙ o is 5
So the answer is 5

As shown in the figure, AB is the chord of circle O, and the radius co od intersects AB at point E F respectively Fig. AB is the chord of circle O. the radius OC and OD intersect AB at points E and f respectively, and AE = BF. Please find out the quantitative relationship between OE and of and give proof The picture is very good, I will not send

OE=OF
Proof: connect Ao, Bo
∵ Ao = Bo, ᙽ OAB is an isosceles triangle
∵ AE = BF
∴△OAE≌△OBF
∴OE=OF

AB is the diameter of circle O, BC is the chord, OD is perpendicular to BC to e, and BC is to d Please write five different types of correct conclusion BC = 8, ed = 2, find the radius

BC ⊥ AC, AC ∥ OD, CE = be, arc CD = arc BD, angle a = angle BOD