As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained

The transition point O is OC ⊥ AB to C, as shown in the following figure:  AOC = 12 ﹤ AOB = 60 °, AC = BC = 12ab, ﹤ in RT △ AOC, ﹤ a = 30 ° OC = 12oa = 10cm, AC = oa2 − oc2 = 202 − 102 = 103 (CM), ab = 2Ac = 203cm ﹥ the area of AOB = 12ab · OC = 12 × 203 × 10 = 1003 (cm2)

As shown in the figure, AB is the diameter of ⊙ o, chord CD ⊥ AB is at point E, tangent of ⊙ o is made through point B, and the extension line of AC is at point F. it is known that OA = 3, AE = 2, (1) Find the length of CD; (2) Find the length of BF

(1) As shown in the figure, connect OC,
∵ AB is the diameter, chord CD ⊥ AB,
∴CE=DE
In the right angle △ OCE, oc2 = oe2 + CE2
32=（3-2）2+CE2
Results: CE = 2
2，
∴CD=4
2．
(2) ∵ BF cut ⊙ o at point B,
∴∠ABF=90°=∠AEC．
And ∵ CAE = ∵ FAB (common angle),
∴△ACE∽△AFB
∴AE
AB=CE
BF
Namely: 2
6=2
Two
BF
∴BF=6
2．

It is known that AB is the diameter of ⊙ o, ⊙ o passes through the midpoint D of BC, and de ⊥ AC Proof: De is tangent of ⊙ o

Proof: connect OD
∵ D is the midpoint of BC, O is the midpoint of ab,
∴OD∥AC，
Ψ CED = ∠ ode. (4 points)
∵DE⊥AC，
﹤ CED = ∠ ode = 90 °. (6 points)
⊥ De, OD is the radius of the circle,
⊙ De is the tangent line of ⊙ O. (10 points)

As shown in the figure, ⊙ o with side ab of △ ABC as diameter passes through the midpoint D of BC and passes through D as de ⊥ AC at E (1) Results: ab = AC; (2) Proof: De is tangent of ⊙ o

Proof: (1) connect ad
∵ AB is the diameter of ⊙ o,
∴AD⊥BC，
And BD = CD,
∴AB=AC．
(2) Connect OD
∵OA=OB，BD=CD，
∴OD∥AC．
And de ⊥ AC,
∴OD⊥DE，
⊙ De is the tangent of ⊙ o

As shown in the figure, in the triangle ABC, the angle ABC = 90 degrees, the circle O with diameter AB intersects AC at D, e is the midpoint of BC. It is proved that De is the tangent line of circle o

It is proved that: the connection BD ∵ AB is the diameter ᙽ ADB = 90 ° [the circumference angle opposite to the angle is the diameter]

As shown in the figure, AB is the diameter of circle O, BC intersects circle O with point D, and De is perpendicular to AC and point E. in order to make de the tangent line of circle O, another condition needs to be added A DE=DO B AB=AC C CD=DB D AC//OD

The incorrect one is a: de = do
The correct condition is B: ab = AC
C：CD=DB
D：AC//OD
The proof is as follows
Connect OD,
1. If the condition B: ab = AC holds, then angle B = angle c; OD = ob, then angle B = angle BDO,
Therefore, if there is an angle c = angle BDO, then od is parallel to AC (indicating that condition D holds), and de ⊥ AC, then de ⊥ OD, so De is the tangent of circle o
2. If D is the midpoint of BC, after connecting OD, OD is the median line of triangle ABC. If od is parallel to AC, de ⊥ AC, then de ⊥ od and de are tangent lines of circle o

It is known that AB is the diameter of ⊙ o, ⊙ o passes through the midpoint D of BC, and de ⊥ AC Proof: De is tangent of ⊙ o

It is proved that: the connection OD ∵ D is the midpoint of BC, O is the midpoint of AB,

As shown in the figure, take the waist AB in the isosceles △ ABC as the diameter to make ⊙ o, the disclosure side BC at point D. make de ⊥ AC through point D, and E (1) Verification: De is tangent line of ⊙ o; (2) If the radius of ⊙ o is 6, ∠ BAC = 60 °, find the length of de

(I) proof: connect od and ad, as shown in Fig,
∵ AB is the diameter of ⊙ o,
⊥ BC is ad ⊥ BC,
∵ △ ABC is an isosceles triangle,
∴DB=DC，
OA = ob,
The OD is the median line of △ ABC,
∴OD∥AC，
∵DE⊥AC，
∴DE⊥OD，
⊙ De is the tangent line of ⊙ o;
（Ⅱ）∵∠BAC=60°，
ν Δ ABC is an equilateral triangle,
∴∠B=∠C=60°，
The △ OBD is an equilateral triangle,
∴BD=OB=6，
∴CD=6，
In RT △ CDE, CE = 1
2CD=3，
∴DE=
3CE=3
3．

As shown in the figure: isosceles △ ABC, take the waist AB as the diameter, make ⊙ o, the bottom edge BC is at P, PE ⊥ AC, and the perpendicular foot is e Verification: PE is tangent of ⊙ o

Proof: connect Op,
∵ AB is the diameter of ⊙ o,
∴∠APB=90°，
∵AB=AC，
∴BP=CP，
∵OB=OA，
∴OP∥AC，
∵PE⊥AC，
∴OP⊥PE，
∵ Po is the radius,
⊙ PE is the tangent of ⊙ o

As shown in the figure, it is known that AB is the chord of ⊙ o, the radius OA = 20cm, ∠ AOB = 120 ° and the area of ⊙ AOB is obtained