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As shown in the figure, AB is the diameter of ⊙ o, ad is the chord, ∠ DBC = ∠ a, OC ⊥ BD at point E (1) It is proved that BC is tangent of ⊙ o; (2) If BD = 12, EC = 10, find the length of AD
(1) Prove that: ∵ AB is the diameter of ⊙ o,
∴∠D=90°，∠A+∠ABD=90°．
∵∠DBC=∠A，
∴∠DBC+∠ABD=90°，
In other words, ABC = 90 °
∴AB⊥BC．
⊙ BC is the tangent line of ⊙ o
（2）∵OC⊥BD，
∴∠OEB=90°，
∴OE∥AD，
∴BE=ED=1
2BD=6．
∵∠BEC=∠D=90°，∠DBC=∠A，
∴△BEC∽△ADB，
∴BE
AD＝EC
DB，
∴6
AD＝10
12．
∴AD=7.2．
As shown in the figure, AB is the diameter of ⊙ o, ad is the chord, ∠ DBC = ∠ a (1) Verification: BC is tangent to ⊙ o; (2) If OC ‖ ad, OC intersects BD at point E, BD = 6, CE = 4, find the length of AD
(1) Prove that: ∵ AB is the diameter,
Ψ d = 90 ° ad ⊥ BD. (1 point)
Ψ a + ∠ abd = 90 °. (2 points)
And ? DBC = ∠ a,
∴∠DBC+∠ABD=90°，
In other words, ABC = 90 °
⊥ BC. (3 points)
∵ ob is the radius,
⊙ BC is tangent to ⊙ O. (4 points)
（2） ∵OC∥AD，∠D=90°，
∴∠OEB=∠D=90°．
/ / OC ⊥ BD. (5 points)
∴BE=DE=1
2bd = 3. (6 points)
∵BE⊥OC，∠OBC=90°，
﹤ OBE ∽ BCE. (7 points)
∴OE
BE=BE
EC is OE
3=3
4，
∴OE=9
4. (9 points)
∵OA=OB，DE=EB，
∴AD=2EO=9
2. (10 points)
As shown in the figure, AB is the diameter of semicircle o, ad is the chord, ∠ DBC = ∠ a, if OC is vertical ad, OC intersects BD with E, BD = 6, CE = 4, find the length of AD
Because OC is parallel to ad, angle cob = angle a, because angle ADB and angle CBO are both right angles, triangle ADB and triangle cob are similar. Let BC intersect ad with F, triangle cob is similar to triangle ABF, because o is the midpoint of AB, so e is also the midpoint of DB
As shown in the right figure, AB is the diameter of circle O, the radius OC ⊥ AB, e is the point on ob, and the chord ad ⊥ CE intersects OC at point F. explore the relationship between line OE and of, and explain the reasons
It is proved that the intersection point G of ad ⊥ CE is set
∵ public ∵ a, OC ⊥ ab
∴△AOF∽△AEG
∴∠AFO=∠CEO
And ∵ AFO = CEO, OC ⊥ AB, OA = OC are the same radius
∴△AOF≌△CEO
∴OE=OF
AB is the diameter of the circle O, the radius OC is perpendicular to AB, e is a point on ob, the chord ad is perpendicular to CE, and OC is at F
When ad intersects CE at point m, the triangle OCE is similar to the triangle ame, and the triangle ame is similar to the triangle AFO (both of which have the same angle and right angle relationship). Therefore, the triangle OCE is similar to the triangle AFO. In the two triangles, OC = OA = the radius of the circle, so the two triangles are congruent, and the corresponding sides of the congruent triangles are equal, so of = OE
AB is the diameter of ⊙ o, OC ⊥ AB, e is a point on ob, the chord ad ⊥ CE intersects OC at point F, and OE = of
Let EC and ad intersect with point G
Because ad ⊥ CE, OC ⊥ ab
As shown in the figure, in the circle O, the radius OC is perpendicular to the diameter AB, e and F are on OA and OC respectively, and OE = of. It is proved that CE ⊥ BF
It was proved that BF was extended to be CE to H
∵OC⊥AB
∴∠COA＝∠COB＝90
∴∠ECO+∠CEO＝90
∵OC＝OB、OE＝OF
∴△CEO≌△BFO （SAS）
∴∠FBO＝∠ECO
∴∠CHB＝∠FBO+∠CEO＝∠ECO+∠CEO＝90
∴CE⊥BF
As shown in the figure, in the circle O, the radius OA is perpendicular to the chord BC, and the perpendicular foot is DOD = 4, ad = 1. Find the lengths of BC and ab
Connect ob
∵OA⊥BC
The vertical diameter theorem: BD = CD = 1 / 2BC
∵OB=OA=AD+OD=1+4=5
∴OB²=BD²+OD²
5²=BD²+4²
Then BD = 3
∴BC=2BD=6
∴AB²=AD²+BD²=1²+3²=10
AB=√10
As shown in the figure, point a is in the circle O, BC is the chord, OD ⊥ BC is in D, where OA = 8, ab = 10, ∠ a = ∠ B = 60 °, then the length of BD is?
Extended Ao intersects BC with E, ⊿ Abe is a regular triangle, OE = 2, OD = √ 3, connected with ob
OB²=100+64-2×10×8÷2=84
BD²=OB²-OD²=84=3=81
BD=9
As shown in the figure, the intersection of AD and BC with the point O, ab ∥ CD, OA = OD, verification: ab = CD As shown in the figure, ab = AC, ∠ 1 = ∠ 2, ∠ 3 = ∠ 4, prove △ Abe ≌ △ ACD
Ab ‖ CD, angle ABC = angle DCB, angle bad = angle ADC, OA = OD
If the triangle AOB is all equal to the triangle doc, then AB = CD
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