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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
It is proved that: connecting OD, as shown in the figure: ∵ OA = OD, ODA = ∵ ad ∵ Co, cod = ∠ ODA, ? cob. cod = ∠ cob. In △ ODC and △ OBC, OD = ob ∠ doc = bococ = OC ≌≌△ OBC (SAS) ODC = ∠ OBC ≌ ≌ CB is tangent of circle O and ob is
AB is the diameter of circle O, and ad is the chord of circle O. the tangent passing through point B intersects with the extension line of ad at point C, and ad = DC?
45 degrees
Just prove that triangle ABC is an isosceles right triangle
It is known that in the circle O, AB is the diameter and ad is the chord. The tangent line passing through point B intersects with the extension line of ad at C, and ad = DC. Find the degree of angle abd
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In the circle whose center is O, AB is the diameter and ad is the tangent line of chord passing through point B. BC and the extension line of ad intersect at point C and ad is equal to CD. What is the degree of angle abd In the circle whose center is O, AB is the diameter, ad is the tangent line of chord passing through point B, BC intersects with the extension line of ad at point C, and ad is equal to the degree of angle abd obtained by CD
Because AB is the diameter of circle O, BC is the tangent of circle o,
So BC is perpendicular to AB, and the angle ABC is 90 degrees,
B is the diameter of circle O, and point D is on circle o,
So the angle ADB is a right angle, BD is perpendicular to AC,
And because ad = CD,
So BD is the vertical bisector of AC,
So Ba = BC
So BD bisection angle ABC,
Because the angle ABC = 90 degrees,
So the angle abd is 45 degrees
In the circle whose center is O, AB is the diameter, ad is the tangent line of chord passing through point B, BC intersects with the extension line of ad at point C, and ad is equal to the degree of angle abd obtained by CD
connect BD.BD If the angle DAB = 45, then the angle abd = 90-45 = 45 degrees
Known: as shown in the figure, in circle O, diameter AB intersects chord CD at point m, and M is the midpoint of CD, point P is on the extension line of DC, PE is the tangent line of circle O, e is the tangent point, AE is the tangent point I can do it already
In the circle O, the diameter AB intersects with the chord CD at point m, and M is the midpoint of CD, point P is the tangent line of circle O, e is the tangent point of circle O, AE and CD intersect at F. it is proved that: ∵ diameter AB and chord CD intersect at point m, and M is the midpoint of CD ? it can be proved that ab ⊥ CD, ? ABF = 90 °, ? AFB + ?
As shown in the figure, the radius OC of ⊙ o is 6cm, and the chord AB bisects OC vertically, then ab=______ cm．
Let the perpendicular foot of AB and OC be point P, connected with OA, as shown in Fig,
∵ chord AB is vertically bisected OC,
∴PA=PB，OP=PC，
The radius OC of ⊙ o is 6cm,
/ / op = 3 and OA = 6,
∴AP=
62−32=3
3，
∴AB=2AP=6
3cm．
So the answer is 6
3．
As shown in the figure, the radius OC of ⊙ o is 6cm, and the chord AB bisects OC vertically, then ab=______ cm，∠AOB=______ .
Let the intersection point of OC and ab be D, as shown in the figure: ∵ radius OC ⊥ AB, ᙽ point D is the midpoint of chord AB, that is, ad = BD = 12ab, and ∵ chord AB vertically bisects OC, and OC = 6cm, od = CD = 12oc = 3cm. In RT △ AOD, OA = OC = 6cm, OD = 3cm
As shown in the figure, the radius OC of ⊙ o is 6cm, and the chord AB bisects OC vertically, then ab=______ cm，∠AOB=______ .
Let the intersection point of OC and ab be D, as shown in the figure:
∵ radius OC ⊥ AB,
The point D is the midpoint of the chord AB, that is, ad = BD = 1
2AB，
And ∵ the chord AB is vertically bisected OC, and OC = 6cm,
∴OD=CD=1
2OC=3cm，
In RT △ AOD, OA = OC = 6cm, OD = 3cm,
According to Pythagorean theorem: ad=
OA2−OD2=3
3cm，
Then AB = 2ad = 6
3cm，
∵OA=OB，OD⊥AB，
ν OC is the bisector of AOB, i.e. ∠ AOC = ∠ BOC = 1
2∠AOB，
In RT △ AOD, sin ∠ AOC = ad
OA=3
Three
6=
Three
2，
∴∠AOC=60°，
Then ∠ AOB = 2 ∠ AOC = 120 °
So the answer is: 6
3；120°
As shown in the figure, AB is the chord (non diameter) of ⊙ o, C and D are two points on AB, and AC = BD Confirmation: OC = OD
It is proved that if O is used as OE ⊥ AB in E, then AE = be, (4 points)
And ∵ AC = BD, ∵ CE = De
The OE is the vertical line of CD, (6 points)
/ / OC = OD. (8 points)
RELATED INFORMATIONS
- 1. As shown in the figure, AB is the chord of ⊙ o, points c and D are on the chord AB, and OC = OD
- 2. 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
- 3. 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
- 4. 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
- 5. 0
- 6. 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 °
- 7. In the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, try to guess the position relationship between AF and CD, and explain the relationship
- 8. 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
- 9. 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．
- 10. 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 |
- 11. 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
- 12. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
- 13. 0
- 14. 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
- 15. In the triangle ABC, the points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and the ad bisection angle BAE (1) when the angle BAC is equal to 90 degrees In the triangle ABC, points D and E are on BC, the angle CAE is equal to angle B, e is the midpoint of DC, BD is equal to AC, and ad bisects the angle BAE (1) When the angle BAC is equal to 90 degrees, it is proved that BD is equal to AC (2) When the angle BAC is not equal to 90 degrees, is there still BD equal to AC? Explain the reason
- 16. As shown in the figure, in the triangle ABC, ad is the bisector of ∠ BAC, EF bisects ad vertically, intersects ad at e, and the extension line intersects BC at F, then ∠ B = ∠ caf? Because ad bisects angle BAC, angle bad = angle CAD Because EF bisects ad vertically, AF = DF, so angle DAF = angle ADF Because angle DAF = angle DAC + angle CAF, angle ADF = angle B + angle bad, angle DAC + angle CAF = angle B + angle bad Because angle bad = angle CAD, angle B = angle caf Why angle ADF = angle B + angle bad
- 17. As shown in the figure, in △ ABC, D is the midpoint of BC, de ⊥ BC, the bisector of ⊥ BAC is AE at point E, EF ⊥ AB is at point F, eg ⊥ AC intersects AC extension at point g. verification: BF = CG
- 18. As shown in the figure, in △ ABC, D is a point on AB, and ad = AC, AE ⊥ CD, the perpendicular foot is e, and F is the midpoint of CB
- 19. As shown in the figure, in diamond ABCD, AE ⊥ BC, perpendicular foot is e, angle DAE = 2, angle BAE, BD = 15, find the length of AC, ab
- 20. As shown in the figure, both the triangle ABC and the triangle Dec are isosceles right triangles. The angle ACB = angle DCE = 90 ° f is the midpoint of De, h is the midpoint of AE, and G is the midpoint of BD If the △ Dec in Fig. 1 is rotated clockwise by an acute angle around point C to Fig. 2, is the triangle FGH still an isosceles right triangle