Elective course 4-1: plane geometry as shown in the figure AB is the diameter of ⊙ o, the extension line of chord BD and Ca intersects at point E, and the extension line of EF perpendicular to Ba intersects at point F. (I) verification: ∠ DEA = ∠ DFA; (II) if ∠ EBA = 30 °, EF = 3, EA = 2Ac, find the length of AF

(I) prove: connect ad, BC, because AB is the diameter of ⊙ o, so ∠ ADB = ∠ ACB = ∠ EFA = 90 °, so four points a, D, e, f are in common circle, and ∩ DEA = ∠ DFA; (II) in right angle △ EFA and right angle △ BCA, so ∩ EFA ∽ BCA, so AFAC = aEAb, so AF × AB = AC ×
As shown in the figure, in the triangle ABC, D is the point on AC, and ad = CD + BC passes through the point D, making the outer circle of AC perpendicular intersecting with m, finding m as the midpoint of superior arc ab
Prove: make the extension line of AC to N, so that CN = BC. Connect BN, am, Mn, MB
∵ ad = DC + BC = DC + CN = DN ∵ DM is the vertical bisector of an ∵ am = Mn
∴∠MAN=∠MNA
And ∵ ∠ MAC = ∠ MBC (because the opposite arcs are equal)
∴∠MNA=∠MBC
And ∵ BC = CN
∴∠CBN=∠CNB
∴∠MBC+∠CBN=∠MNA+∠CNB
Ψ MB = Mn ∵ Ma = Mn ∵ Ma = MB
Ψ arc Ma = arc MB
It is known that the radius of circle O is 2. Take the chord Ba of circle O as the diameter to make circle M. point C is a moving point on the superior arc ACB of circle O (not coincident with points B and a). Connect Ca and BC, intersect with circle m respectively, and connect de with points D and E, if Ba = 2 times the root sign. 3. Calculate the degree of angle c and the length of de
C is 60 ° De, and the length is the root sign. 3. Draw the general figure first, and according to the relevant values, the center angle AOB is 120 ° and the circumference angle c is half of the center angle. Connect AE and BD, and the angle ADB and angle AEB in the circle M are 90 °. In the right triangle ace, the angle c is 60 ° and CE = 1 / 2Ac, similarly in the triangle BCD, CD = 1 / 2BC, in the triangle CDE, COSC = (CD ^ 2 + CE ^ 2-DE ^ 2) \ (CD * ce). In the triangle ABC, the angle c is 60 ° and CE = 1 / 2ac, COSC = (AC ^ 2 + BC ^ 2-AB ^ 2) \ (AC * BC), substituting into the correlation formula and relevant values, you can get 4de ^ 2 = AB ^ 2, you can get de = 1 / 2Ab
Circle P and circle O intersect at two points a and B. circle P passes through center O. point C is any point on superior arc AB of circle P, connecting AB, AC, BC and OC
Do you want these answers? (1) point out an angle that is equal to the angle ACO in the graph; ∠ ACO = ∠ BCO (2) when the point C is in the circle P, the straight line CA is tangent to the circle O? Explain the reason. When the point C is in the circle O, the straight line CA is tangent to the circle O, connects OP and extends, and the intersection circle O connects AD and OA at the point D
(1) Connect OA, ob
In ⊙ o, ∵ OA = ob,
Qi
OA
=
OB
,
∴∠ACO=∠BCO；
(2) Connect OP and extend the intersection point d with ⊙ P
If point C is at point D, the line CA is tangent to ⊙ o
Reason: connect AD and OA, then ∠ Dao = 90 degree
∴OA⊥DA
⊙ Da is tangent to ⊙ o
... unfold
(1) Connect OA, ob
In ⊙ o, ∵ OA = ob,
Qi
OA
=
OB
,
∴∠ACO=∠BCO；
(2) Connect OP and extend the intersection point d with ⊙ P
If point C is at point D, the line CA is tangent to ⊙ o
Reason: connect AD and OA, then ∠ Dao = 90 degree
∴OA⊥DA
⊙ Da is tangent to ⊙ o
When point C is at point D, the line CA is tangent to ⊙ o
(3) When ∠ ACB = 60 ° the radii of the two circles are equal;
Reason: make diameter OD, connect BD, ad, OA,
∵∠ ADB = ∠ ACB = 60 ° Po vertical bisection AB,
Qi
AO
=
BO
,
∵∠ADO=∠BDO，
∴∠ADO=30°，
∵ od is the diameter,
∴∠DAO=90°，
∴OA=
One
Two
OD，
∴OA=PO，
When ∠ ACB = 60 ° the radii of the two circles are equal
The radius of circle O is known to be 2. Take the chord ab of circle O as the diameter to make a circle M. thank you, great God for your help
Given that the radius of circle O is 2, take the chord ab of circle O as the diameter of circle m, if AB = 2 times root 3, point C is a moving point (not coincident with point a and point B) on the superior arc AB of circle O, connect AC and BC, intersect with circle m at point D and point e respectively, connect de. 1, find the degree of C. 2, find the length of de. 3, if you remember tanabc = y, AD / DC = x (x greater than 0, less than 3), then in the process of point C's movement, try to use the algebraic expression containing x to express y
In the triangle AOB, oh is perpendicular to AB, and the perpendicular foot is h. according to the vertical diameter chord splitting theorem, ah = BH = 0.5ab = root 3, and because OA is the radius and OA = 2, the angle AOH = 60 degrees is obtained in the right triangle AOH; similarly, the angle boh = 60 degrees, so the angle AOB = 120 degrees, because the angle c is the inner angle of the circle opposite to the inferior arc AB, and the angle AOB is the center angle opposite to the inferior arc AB; so the angle c = 0.5, and the angle AOB = 60 degrees
It is known that: as shown in the figure, AB is the diameter of ⊙ o, AC is the chord of ⊙ o, M is the point on AB, passing through point m to make DM ⊥ AB, intersecting chord AC at point E, intersecting ⊙ o at point F, and DC = de. (1) prove that DC is the tangent of ⊙ o; (2) if DM = 15, CE = 10, cos ∠ AEM = 513, find the length of radius of ⊙ o
(1) As shown in the figure, connect OC, \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\d The radius of circle O is 12ab = 24730 when m = 5, EM = dm-de = 2, ame = DGE = 90, AEM = DEG, AMDG = emeg = Aede, namely am12 = 25 = ae13, am = 245, AE = 265, AC = AE + EC = 765, AB is the diameter of circle O, ACB = 90, cosa = amae = acab, ab = 24715
As shown in the figure, AB is the diameter of ⊙ o, the chord CD ⊥ AB is at point m, passing through point B to be ∥ CD, the extension line of AC is at point E, connecting BC. (1) prove that be is the tangent of ⊙ o; (2) if CD = 6, BM: cm = 1:2, find the diameter of ⊙ o
(1) It is proved that: ∵ be ∥ CD, ab ⊥ CD, ∵ ab ⊥ be. ∵ AB is the diameter of ⊙ o, and ∵ be is the tangent of ⊙ O. (2) ∵ AB is the diameter of ⊙ o, ab ⊥ CD,

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