As shown in the figure, circle O and circle 1 intersect at points a and B, AC is the diameter of circle O, and the extension lines of Ca and CB intersect circle O1 at points D and E, AC = 12, be = 30, BC = ad, Find ∠ C and de

As shown in the figure, circle O and circle 1 intersect at points a and B, AC is the diameter of circle O, and the extension lines of Ca and CB intersect circle O1 at points D and E, AC = 12, be = 30, BC = ad, Find ∠ C and de

What about the picture?
Are the two centers on one side
Prompt
Connect AE
Let CB be X
Angle ABC is a right angle
Use similar triangles yourself

It is known that circle O1 and circle O2 intersect at points a, B, and pass through point B as CD, perpendicular to AB, and intersect circle O1 and circle O2 at C, D, respectively, When the straight line EF rotates around point B, whether the ratio of AE to AF remains unchanged? (2) if AB = 2, CD = 6, when the line EF rotates around point B, whether the maximum AEF of trigonometry is obtained. If there is any, ask for and explain the reason. If not, please explain the reason

To solve this problem, first of all, we must explain that the chord AC is the diameter of two circles (1) is to prove AE: AF = AC: ad. methods to prove that the triangle ace and triangle ADF (2) should make full use of intuition, it is easy to find that the area of triangle AEF is ACD

Given that the circle O1 and O2 intersect at point a, B passes through point B to make CD vertical AB, respectively intersect circle O1 and circle O2 at point C, D and any straight line passing through point B to intersect circle O1 and circle O1 respectively It is proved that AC and AD are the diameters of circle O1 and circle O2 at points E and f respectively

It is proved that: ∵ ab ⊥ CD  ABC ∵ ADC = 90 ᙽ AC, ad is the diameter of circle O1 and circle O2 ? AEB, and the arcs corresponding to ? ACB ∵ ADB and ? AFB are all inferior arcs, ab ᚉ ADB ∵ AFB ∵ AEF is similar to △ ACD ∵ AE / AF = AC / ad ∵ AC, AD are circles

It is known that: as shown in the figure, circle 1 and circle 2 intersect at two points a and B, C is a point on circle O1, AC intersects circle O2 at point D, and straight line EF crosses O1, O2 at e, F. try to explain EC ∥ D Verification: EC parallel DF

prove:
Connect ab
∵ quadrilateral ABEC inscribed in circle O1
∴∠ABF=∠C
∵ the quadrilateral ABFD is inscribed in circle O2
∴∠ABF+∠D=180°
∴∠C+∠D=180°
∴CE‖DF

As shown in the figure, it is known that circle O1 and circle O2 intersect at points a and B. the straight lines CD and EF passing through point a intersect circles O1 at D and F, and circles O2 at C and e respectively. It is proved that: ∠ CBE = ∠ DBF

It is proved that in ⊙ O1, arc AB = arc ab
∴∠BFA=∠BDA
In ⊙ O2, arc AB = arc ab
∴∠AEB=∠ACB
∴∠FBE=∠DBC
∴∠FBE-∠DBE=∠DBC-∠DBE
∴∠DBF=∠CBE

It is known that: as shown in the figure, ⊙ O1 and ⊙ O2 intersect at point a and point B, and point O1 is on ⊙ O2, the line CD passing through point a intersects with ⊙ O1 and ⊙ O2 at points c and D, and the line EF passing through point B intersects with ⊙ O1, ⊙ O2 at points E and F, and chord o1d of ⊙ O2 crosses AB at P Verification: (1) ce ∥ DF; （2）O1A2=O1P•O1D．

It is proved that: (1) ∵ quadrilateral ABEC is an inscribed quadrilateral of ⊙ O1,  Abe +  C = 180 °. The quadrilateral ABFD is an inscribed quadrilateral of ⊙ O2,

AB is the diameter of circle O, point D is on circle O, BC is tangent line of circle O, ad ∥ OC

Connect OD,
∵ AB is the diameter of circle O and BC is the tangent of circle o
∴∠CBO=90°
∵OD=OB,CD=CB,OC=OC
∴△COD≌△COB
∴∠CDO=∠CBO=90°
/ / CD is the tangent of circle o

As shown in the figure, OC ⊥ OA intersects circle O at point B, e is a point on circle O, AE intersects OC at point D, and CE = CD Hurry up ~! Please answer carefully I add points

Proof: connect OE
CE = CD, then ∠ CED = ∠ CDE;
Also, it has an option of CDE= options ADO. therefore, an option of AED= options ADO;
If OE = OA, then ∠ OEA = ∠ oad
OC vertical OA, then ∠ ADO + oad = 90 degrees
Therefore, ∠ AED + ∠ OEA = 90 degrees (equivalent substitution)
CE is tangent to circle o

As shown in the figure, AB is the diameter of circle O. the tangent line of circle O is made through point B, and the point on the tangent line C is the point on the tangent line. The extension line connecting OC to circle O to e and AE to BC to d If AB = BC = 2, find the length of CD

OB=1,BC=2
Then OC = √ 5
∴CE=√5-1
∵∠CED=∠AEO=∠A=∠CBE,∠C=∠C
∴△CED∽△CBE
∴CE²=CD*CB
That is (√ 5-1) 2 = 2CD
∴CD=3-√5

As shown in the figure, AB is the diameter of ⊙ o, the radius OC ⊥ AB, D is a point on the extension line of AB, and D is the tangent line of ⊙ o, and E is the tangent point, connecting CE and crossing AB at point F (1) Results: de = DF (2) With AE, if of = 1, BF = 3, find the length of de

(1) Connect OE,
∵ De is the tangent line of the circle,
∴OE⊥ED，
∴∠OEC+∠CED=90°，
∵OC⊥AD，
∴∠COD=90°，
∴∠C+∠CFO=90°，
∵∠CFO=∠DFE，
∴∠C+∠DFE=90°，
∵OC=OE，
∴∠C=∠OEC，
∴∠DFE=∠DEF，
∴DE=DF；
(2) In RT △ OED, OE = ob = of + FB = 1 + 3 = 4,
According to Pythagorean theorem, OD2 = oe2 + ED2, i.e. (1 + DF) 2 = (1 + de) 2 = 42 + de2,
The result is: de = 7.5