AB is the diameter of ⊙ o, CD, CB are the tangent lines of ⊙ o, B and D are tangent points, connecting OC to ⊙ o to e, straight line AE to BC, BD to F, G. verification: AF bisection ⊙ bad

AB is the diameter of ⊙ o, CD, CB are the tangent lines of ⊙ o, B and D are tangent points, connecting OC to ⊙ o to e, straight line AE to BC, BD to F, G. verification: AF bisection ⊙ bad

According to the tangent length theorem: angle doc = angle BOC. So Arc de = arc be. So angle DAF = angle BAF

As shown in the figure, AB is the diameter of ⊙ o, D is the midpoint of arc BC, the extension line of de ⊥ AC crossing AC is at e, the tangent of ⊙ o is the extension line of ad at F (1) It is proved that De is tangent of ⊙ o; (2) If de = 3, the radius of ⊙ o is 5

(1) ⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙⊙

Definition: tangent angle: the angle where the vertex is on a circle, one side intersects with the circle, and the other side is tangent to the circle Problem scenario: as shown in the figure, the straight line AB is the tangent line of ⊙ o, the tangent point is C, CD is a chord of ⊙ o, and ∠ P is the circular angle of arc CD (1) Conjecture: the relationship between tangent angle ∠ DCB and ∠ P. try to prove your conjecture by connecting CO and extending intersection ⊙ o at point E and connecting De (2) State in your own words the conclusion you have guessed

（1）∠DCB=∠P；
It is proved that ∵ CE is the diameter of ⊙ o,
∴∠DCE+∠E=∠EDC=90°；
And ∵ AB is the tangent of ⊙ o,
∴∠DCE+∠DCB=90°，
∴∠DCB=∠E；
And ∵ e = ∵ P,
∴∠DCB=∠P．
(2) The tangent angle is equal to the circular angle of the arc pair on both sides
(or the degree of the tangent angle is equal to half of the degree of the arc between its two sides.)

It is known that: as shown in the figure, take the oblique edge ab of RT △ ABC as the diameter to make ⊙ o, and D is the point on ⊙ o, and there is AC = CD. Pass through point C to make the tangent of ⊙ o, and connect CD with the extension line of BD at point E (1) Try to judge whether be and CE are perpendicular to each other, please explain the reason; (2) If CD = 2 5，tan∠DCE=1 2. Find the radius length of ⊙ o

(1) ∵ AB is the diameter
∴∠ACB=90°
∵AC=CD，
∴∠ABC=∠CBE，
∵ CE is the tangent of ⊙ o,
∴∠BCE=∠A，
∴∠BEC=∠ACB=90°
∴BE⊥CE．
(2) ∵ CE is tangent, AC = CD,
∴∠DCE=∠DBC=∠ABC，tan∠DCE=1
Two
∴tan∠ABC=1
Two
∵AC=CD=2
Five
∴BC=4
Five
∴AB=10
The radius of ⊙ o is equal to 5

B. C is the point on circle O. the segment AB passes through the center O and connects AB and BC. The point C is CD ⊥ AB in D, ﹤ ACD = 2 ∠ B. is AC the tangent of circle O? Why?

Yes
prove:
AB intersects circle O at another point E, be is the diameter, connecting CE, then ∠ ECB = 90 °
∴△BEC ∽ △CED （∠CDE = ∠ECB = 90°,∠CED = ∠BEC）
∴∠B = ∠ECD
And ∵ ACD = 2 ∵ B
∴∠ACE = ∠B
Ψ ace is the tangent angle
/ / AC is the tangent of circle o

As shown in the figure, line AB passes through point C on ⊙ o, and is bisected by point C. OA and ob are respectively delivered to D and e of ⊙ o, ad = be. It is proved that AB is the tangent of ⊙ o

prove:
Because ad = be do = EO
All Ao = Bo
And because point C bisects ab
So AC = BC
So triangle ACO and triangle BCO are congruent
So angle ACO = angle BCO = 90 degrees
So co ⊥ ab
So AB is the tangent of circle o
I haven't done geometry for a long time. I don't know if it's right. Do you see

As shown in Figure 1, ad is the diameter of the center O, B and C are the two points on the center O, point C is on the arc AB, and the arc AB = arc CD, the tangent line of the center O is made through point a, Intersect BD with the extension line at e, and make the vertical line of DC through e. the vertical foot is f (1) Verification ∠ AED = ∠ ADF (2) Explore the relationship among BD, be and EF, and prove that (3) As shown in Figure 2, if point B is on arc AC, other conditions remain unchanged. When AE = 6 and radius o of circle center is 4, find the length of EF (answer with the knowledge of the third semester of junior high school)

(1) If we connect AC because the arc AB = arc CD, then AB = CD, then ∠ ADB = ∠ DAC (equal chords corresponding to the center angle of the circle are equal) because ∠ ADB = ∠ DAC, ∠ DBA = ∠ ACD = 90 degrees (diameter diagonal is 90 degrees), ad = ad, then triangle DBA is congruent triangle ACD, then ∠ DAB = ∠ ADC, because AE is tangent, ∠ DAE = 90 degrees, then ∠ AED = 90 degrees - ∠ ade

It is proved that the angle of intersection of a C and a C is equal to the diameter of a circle

prove:
∵ AB is the diameter, ∵ cab = 30 °
∴∠ACB=90º,∠CBA=60º
∵ CP is tangent
[tangent angle of chord is equal to the circular angle of the arc clamped to] PCB= [CAB=30]
∵∠P=∠CBA-∠PCB=60º-30º=30º
∴∠P=∠CAB
∴AC=CP

As shown in the figure, in ⊙ o, M is the midpoint of the chord AB, passing through point B as the tangent of ⊙ o, and intersecting with OM extension line at point C (1) It is proved that: ∠ a = ∠ C; (2) If OA = 5, ab = 8, find the length of OC

As shown in the figure on the right, (1) it is proved that the connection ob, ∵ BC is tangent,  OBC = 90 °,  OBM +  CBM = 90 °, ? OA = ob,  a = ∠ OBM, ? m is the midpoint of AB, ? om ⊥ ab.  C + ∵ CBM = 90 °,  C = ∠ OBM, ∠ a = ∠ C; (2) ? C = ∠ OBM, ∠ OMB = 90

As shown in the figure, a is a point outside the circle O with radius 1, OA = 2, AB is the tangent line of ⊙ o, B is the tangent point, the chord BC ∥ OA, connecting AC, then the area of the shadow part is equal to () A. Three Four B. π Six C. π 6+ Three Eight D. π 4− Three Eight

Connect ob, OC,
∵ AB is the tangent line of the circle,
∴∠ABO=90°，
In the right angle △ ABO, OB = 1, OA = 2,
∴∠OAB=30°，∠AOB=60°，
∵OA∥BC，
Ψ cob = ∠ AOB = 60 ° and s shadow = s △ BOC,
△ BOC is an equilateral triangle with a side length of 1,
The shaded part of S = s △ BOC = 1
2×1×
Three
2=
Three
4．
Therefore, a