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AB is the diameter of the circle, the radius OC ⊥ AB, the chord EF is parallel AB through the midpoint D of OC. It is proved that the angle Abe is equal to 15 ° and the angle CBE is equal to 30 °
Because OC is perpendicular to AB, EF / / AB, OC is perpendicular to EF, so the angle emo = 90 degrees because om = 1 / 2oC = 1 / 2oe, so the angle MEO = 30 degrees, because EF / / AB, the angle AOE = angle MEO = 30 degrees, because OC is perpendicular to AB, the angle AOC = 90 degrees, because the angle AOE = 30 degrees, the angle EOC = 60 degrees, because the angle Abe = 1 / 2 angle AOE
Make a circle, AB is the diameter of circle O, the radius CO is perpendicular to AB, and D is the midpoint of OC. Make chord EF parallel to ab through D, and calculate the degree of angle CFD Self made circle
30 degrees, if you connect of, you can see that the triangle COF is an equilateral triangle! DF is the vertical line of CO! So it's 30 degrees!
Given that ABCDEF is a regular hexagon, and vector AB = a, vector AE = B, use a, B to represent vector BC [need to have a solution process]
Hexagons were proved to be ad parallel BC and BC = ad / 2 ab parallel De
Vector de = vector AB = a
Vector ad = a + B
Vector BC = 1 / 2 (a + b)
As shown in the figure below, in the regular hexagon ABCDEF, if vector AB = a, vector AF = B, use vector a, B to express vector AC, vector ad, vector AE Please draw a picture by yourself. Thank you!
AC=2a+b
AD=2a+2b
AE=a+2b
It is known that ABCDEF is a regular hexagon, and → AB=a, → AE=b, using a and B to represent vector → DE=_______
Vector → DE=- a
In the regular hexagon ABCDEF, if the vector AE is m and the vector ad is n, then the vector Ba is
DE=AE-AD=m-n
BA=DE=m-n
If the side length of a regular hexagon ABCDEF is 1, then the vector ab · vector (CB + BA)= Regular hexagonal letter order counterclockwise from left to right dcbafe
0
0
It's equal to the vector CB = - 1 / 3 (a + b)
In the regular hexagon ABCDEF with side length 1, then the vector ac * vector BD=
You draw a sketch map by yourself, vector BD = vector AE, triangle ace is a regular triangle, so the angle between the two vectors is 60 ° because it is a regular hexagon, so the module of vector BD is root 3, so vector ac * vector b = the square of root sign three, and then multiply cos60 ° = 1.5
It is known that the hexagon ABCDEF is a positive hexagon, the tangent vector AC is equal to the vector a, and the vector BD is equal to the vector B, which are respectively represented by ab Denotes the vector de ad BC EF FA CD AB CE
A point is the origin to establish a rectangular coordinate system, a (0,0), B (1 / 2, - SQR (3) / 2), C (3 / 2, - SQR (3) / 2), C (3 / 2, - SQR (3) / 2), D (2,0), e (3 / 2, SQR (3) / 2), f (1 / 2, SQR (3) / 2), a = AC = (3 / 2, - SQR (3) / 2) B = BD = D = b = (3 / 2, SQR (3) / 2) = b = BD = D = D-B = (3 / 2, SQR (3) / 2) = Aede = ae-ad = 1 / 3b-2 / 3aad = 2 / 3 (a + b) BC = 1 / 2ad = 1 / 3 (A / 3) a (a) a (a + b) B = 1 / 3+ b) EF = - B
RELATED INFORMATIONS
- 1. In the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, try to guess the position relationship between AF and CD, and explain the relationship
- 2. Two equal line segments AB and CD coincide in one third. M and N are the midpoint of AB and CD respectively. If Mn = 12cm, find the length of ab
- 3. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 120 ° and EF is the vertical bisector of ab. EF intersects BC at point F and ab at point E. verification: BF = 1 2FC．
- 4. It is known that in △ ABC, CA = CB, ∠ C = 90 ° D is any point on AB, AE ⊥ CD, perpendicular foot is e, BF ⊥ CD, and vertical foot is F. it is proved that EF = | ae-bf |
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- 6. As shown in FIG. 12, AOB is a straight line, OE bisection of AOC, of bisection of BOD, ∠ cod = 30 ° and try to find the degree of ∠ EOF
- 7. The surface area of regular hexagon is 12 If it is divided into six regular triangles, Is an equilateral triangle?
- 8. The ratio of the three sides of a triangle is 2:4:5. The longest side is 6cm longer than the shortest side. Find out the circumference of the triangle with the equation
- 9. If an equilateral triangle is a meter long, its circumference is () meters
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- 12. As shown in the figure, the radius of ⊙ o is known to be 1, De is the diameter of ⊙ o, the tangent of ⊙ o is ad, C is the midpoint of AD, AE intersects ⊙ o at point B, and the quadrilateral bcoe is a parallelogram (1) Find the length of AD; (2) Is BC tangent to ⊙ o? If so, give proof; if not, give reasons
- 13. As shown in the graph, points D, e and F are the points on the edges BC, Ca and ab of the triangle ABC, De is parallel to Ba, DF is parallel to Ca, and it is proved that ∠ FDE = ∠ a
- 14. As shown in the figure, it is known that AB is parallel to CD, OA = OD, BC passes through point O, e, f are on the straight line AOD and AE = DF
- 15. As shown in the figure, AB is the chord of ⊙ o, points c and D are on the chord AB, and OC = OD
- 16. It is known that AB is the diameter of ⊙ O. the straight line BC and ⊙ o are tangent to point B. through a, ad ⊙ OC intersects ⊙ o at point D and connects CD Proof: CD is tangent of ⊙ o
- 17. It is known that AC and BD are two mutually perpendicular chords of the circle O: x2 + y2 = 4, and the perpendicular foot is m (1, 2) The maximum area of the quadrilateral ABCD is () A. 4 B. 4 Two C. 5 D. 5 Two
- 18. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
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- 20. As shown in the figure, in △ abd and △ ace, ab = ad, AC = AE, ∠ bad = ∠ CAE, connecting BC and de intersecting at point F, BC and ad intersecting at point g. verification: BC = De