0

0

It is proved that OA = OD, OB = OC, angle AOB = angle doc
Because the two triangles are congruent, the angle a = angle D and the angle is staggered, so AB is parallel to DC

As shown in Fig. AB, tangent circle O is at point B, OC is vertical to OA, and OA intersects point D at BC, so as to find the size relationship between AB and AD

AB=AD
Because the angle abd = angle ABO - angle OBC = 90 ° - angle OBC;
Angle ADB= angle ODC (opposite top angle)
Angle ODC = 90 ° - angle OCB (angle AOC is right angle)
Angle OCB = angle OBC;
Therefore, the 90 ° - angle OBC = 90 ° - angle OCB
That is: angle abd = angle ODC = angle ADB
So the triangle abd is an isosceles triangle
AB=AD

It is known that: as shown in the figure, ad and BC intersect at point O, OA = OD, OB = OC （1）△ABO≌△DCO；  （2）AB∥CD．

It is proved that in △ AOB and △ cod, the,
In △ ABO and △ DCO,
OA＝OD
∠AOB＝∠DOC
OB＝OC ，
∴△AOB≌△COD（SAS）；
（2）∵△AOB≌△COD，
∴∠A=∠D，
∴AB∥DC．

AB is the diameter of circle O, BC ⊥ AB, DC is the tangent of circle O. if the radius is 2, the value of ad × OC is

Connect BD, OD, OC
∵ CD is the tangent of circle O, BC ⊥ ab
∴∠CDO＝∠CBA＝90
∵OB＝OD,OC＝OC
∴△BOC≌△DOC （HL）
∴∠BOC＝∠DOC＝∠BOD/2
∵OA＝OD
∴∠BAD＝∠ODA
∴∠BOD＝∠BAD+∠ODA＝2∠BAD
∴∠BOC＝∠BAD
∵ diameter ab
∴∠ADB＝∠CBA＝90
∴△ABD∽△COB
∴AD/AB＝OB/OC
∴AD×OC＝AB×OB＝4×2＝8
The mathematics tutor group has answered your question. Please accept it as the best answer

As shown in the figure, AB is known to be the diameter of ⊙ o, CD is the chord, and ab ⊥ CD is connected to AC, OC, BC at point E (1) It was proved that ∠ ACO = ∠ BCD; (2) If EB = 8cm, CD = 24cm, find the diameter of ⊙ o

(1) It is proved that: connecting OC, ∵ AB is the diameter of ⊙ o, CD is a chord, and ab ⊥ CD is e,  CE = ed, CB = dB. (2 points)  BCD = ∠ BAC. (3 points) ∵ OA = OC,

As shown in the figure, it is known that AB is the diameter of circle O and CD is a chord, and ab ⊥ CD is connected with AC, DC, BC at point E, and ∠ ACO = ∠ BCD I can't draw the picture. Can you imagine it

∠BCD=∠BDC
∠BDC=∠CAB
∠BCD=∠CAB
∠CAB=∠ACO
∠ACO=∠BCD

As shown in the figure, AB is the diameter of circle O, CD is the chord, and ab is perpendicular to CD at point E, connecting acoc, BC (2) Find the diameter of circle O

∵ AB is the diameter of circle O, CD is the chord, and ab is perpendicular to point E, ᙽ CE = de = 1 / 2CD = 1 ᙽ cm
Let the radius of the circle be r, ∵ ab ⊥ CD ᙽ OEC be a right triangle and ∵ CEO = 90 °
 1  2 + (8-r) 2 = R  according to the triangle Pythagorean theorem)
∴R=65/16㎝
The diameter of circle O is 65 / 8 cm

The diameter ab of circle O is perpendicular to CD, the chord EF is perpendicular to OC, and the angle EBC is 2 times of angle Abe

1. Let EF, OC intersect with G and connect OE, because angle AOE is the circumference angle of angle Abe opposite to arc, so angle AOE = 2 * angle Abe is the same as angle COE = 2 * angle CBE because EF is vertically bisecting OC and OE, OC is radius, so og = 1 / 2oC = 1 / 2oe, so angle COE = 60 degrees, because AB is vertical CD, angle COE + angle AOE = 90 degrees, so angle AOE = 30

As shown in the figure, AB is the diameter of semicircle, OC is perpendicular to AB, D is the midpoint of OC, chord D is made through point D, EF is parallel to AB, it is proved that angle Abe = 1 / 2 angle EBC

Connect OE, OC ⊥ AB, D is the midpoint of OC, EF ∥ AB,  Edo = 90 °
In RT ⊿ OED, OC = (1 / 2) OE, then ∠ OED = ∠ 30 ° and ∠ EOD = ∠ 60 °
∴∠ABE=(1/2)∠AOB,∠EBC=(1/2)∠EOC,
∴∠ABE=(1/2)∠EBC

It is known that: as shown in the figure, AB is the diameter of circle O, the radius OC is perpendicular to AB, M is the midpoint of OC, and the chord EF of circle O passes through point m and is parallel to ab. verification: angle CBE = 2 angle Abe

Connect OE, OM = OC / 2 = OE / 2, OC is perpendicular to AB, angle OEM = 30 degrees. EF / / AB, angle AOE = angle OEM = 30 degrees. [internal stagger angle] angle EOC = 90 degrees - angle OEM = 90 degrees - 30 degrees = 60 degrees. Angle CBE = angle EOC / 2 = 30 degrees, [same arc circle angle = circle center angle / 2] angle Abe = angle AOE / 2 = 15 degrees, [same arc circumference angle = circle center angle / 2] angle CBE = 2 angle Abe