It is known that, as shown in the following four graphs, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects with circle O1 at C, intersects with circle O2 at D, and straight line ef? 2 passing through point B As shown in the following figure, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects with circle O1 at C, and with circle O2 at D. the line EF passing through point B intersects with circle O1 at e and circle 02 at F, so as to verify CE ∥ DF Picture in space photo album

It is known that, as shown in the following four graphs, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects with circle O1 at C, intersects with circle O2 at D, and straight line ef? 2 passing through point B As shown in the following figure, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects with circle O1 at C, and with circle O2 at D. the line EF passing through point B intersects with circle O1 at e and circle 02 at F, so as to verify CE ∥ DF Picture in space photo album

No! Because circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects circle O1 at C, and circle O2 at D. the line EF passing through point B intersects circle O1 at e and circle 02 at F. therefore, CE \ \ DF is wrong

As shown in the figure, circle O2 intersects two points a and B, point O1 is on circle O2, the central line of two circles intersects circle O1 at points E and D, circle O2 at F, and ab at C. please refer to the figure Write the relationship between the two line segments

ED×AC=EA×AD
AO1=B01
AE=BE
FA×AO1=FO1×AC

As shown in the figure, we know that AB is the diameter of circle O, make tangent line of circle O through point B, take any point C at tangent line and connect Co. if AD / / OC, prove that CD is tangent line of circle o

prove:
∵AD//OC
﹤ cob = ∠ Dao [equal position angle]
∠ cod = ∠ ODA
∵OA=OD
∴∠DAO=∠ODA
∴∠COB=∠COD
And ∵ ob = OD, OC = OC
∴⊿COB≌⊿COD（SAS）
∴∠CDO=∠CBO
∵ BC is tangent
∴∠CBO=90º
∴∠CDO=90º
/ / CD is the tangent of circle o

As shown in the figure, in circle O, AB is the diameter, BC and circle O are tangent to point B, connect Co, AD are parallel to OC and intersect circle O at point D. It is proved that CD is the tangent of circle o

Because AD / / OC, be / de = Bo / AO = 1, so e is the midpoint of BD. because the triangle BDO is an isosceles triangle, OC is perpendicular to BD. even if OC is the vertical center line of BD, CB = BD, so the triangle BCO is equal to DCO, so the angle CDO = CBO = 90 degrees, so CD is tangent

As shown in the figure, AB is the diameter of ⊙ o, BC ⊥ AB is at point B, OC is connected with ⊙ o at point E, DE= Verification: （1）AD∥OC； (2) CD is the tangent of ⊙ o

Proof: connect OD
（1）∵
DE=
BE，
Ψ DOE = ∠ BOE
∴∠COB=1
2∠DOB．
∵∠DAO=1
2 ∠ DOB (the circle angle opposite to the same arc is half of the center angle of the circle to which it is opposite),
ν (COB = ∠),
∥ OC (the same position angle, two lines are parallel);
（2）∵BC⊥AB，
Ψ CBA = 90 °, i.e. ∠ CBO = 90 °
In △ doc and △ BOC,
DO＝BO
∠DOC＝∠BOC
OC＝OC ，
Then △ doc ≌ △ BOC (SAS) is used,
⊙ CDO = ∠ CBO = 90 ° i.e., CD is tangent of ⊙ o

As shown in the figure, AB is the diameter of ⊙ o, the tangent BC of ⊙ o is made through point B, OC intersects ⊙ o at point E, and the extension line of AE intersects BC at point D (1) It was proved that CE2 = CD · CB; (2) If AB = BC = 2cm, find the length of CE and CD

It is proved that: (1) the connection of be ∵ B ∵ BC is ⊙ the tangent of ⊙ o ∵ B ∵ B ⊙ a ⊙ a ⊙ a ⊙ a ⊙ a ⊙ a ⊙ a ⊙ a ⊙ a ⊙ B ? B ? B ∵ B ∵ B ∵ B ? a ∵ B ᙽ a ⊙ a ⊙ a ⊙ a ⊙ a 9 a ⊙ a CBE ν CECB

AB is the diameter of circle O, BC is perpendicular to B, connected with OC, passing a is ad parallel OC, passing a is ad parallel OC, crossing circle O is D, proving that CD is tangent of circle o

To connect OD, you only need to verify OD ⊥ CD
Because of AD / / OC, ∠ Dao = ∠ cob, ∠ ADO = ∠ cod
Because OA = OD, ∠ Dao = ∠ ADO, then ∠ cob = ∠ cod
Because od = ob, OC is a common edge, then △ OCD and △ OBC are congruent, so ∠ ODC = ∠ OBC
Because BC is perpendicular to B, so ∠ OBC = 90 ° then ∠ ODC = 90 °, that is, OD ⊥ CD
Then CD is the tangent of circle o

AB is the diameter of circle O, D is the midpoint of arc BC, De is perpendicular to AC, AC extension line is at e, tangent BF of circle O intersects extension line of ad at F, if De is 3, radius of circle O is 5, find DF

Let d be the midpoint of the arc BC

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) It is proved that: connect OD, BC, where OD and BC intersect at point G, ∵ D is the midpoint of arc BC, ∵ AB is the diameter of ⊙ o,

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) It is proved that OD, BC are connected, and OD and BC intersect at point G,
∵ D is the midpoint of the arc BC,
ν od vertical bisection BC,
∵ AB is the diameter of ⊙ o,
∴AC⊥BC，
∴OD∥AE．
∵DE⊥AC，
∴OD⊥DE，
∵ od is the radius of ⊙ o,
⊙ De is the tangent of ⊙ o
(2) According to (1), OD ⊥ BC, AC ⊥ BC, de ⊥ AC,
The quadrilateral decg is rectangular,
∴CG=DE=3，
∴BC=6．
⊙ the radius of O is 5,
∴AB=10，
∴AC=
AB2−BC2=8，
It is known from (1) that De is the tangent of ⊙ o,
ν de2 = EC · EA, i.e. 32 = (ea-8) ea,
AE = 9
∵ D is the midpoint of arc BC,
∴∠EAD=∠FAB，
∵ BF cut ⊙ o in B,
∴∠FBA=90°．
And ∵ de ⊥ AC to E,
∴∠E=90°，
∴∠FBA=∠E，
∴△AED∽△ABF，
∴BF
DE＝AB
AE，
∴BF
3＝10
9，
∴BF=10
3．