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As shown in the figure, AB is the chord of ⊙ o, points c and D are on the chord AB, and OC = OD
It is proved that the point O is oh ⊥ AB, and the vertical foot is h, (1 point)
ν ah = BH, (2 points)
∵ OC = OD, and oh ⊥ CD,
ν ch = DH, (4 points)
∴AH-CH=BH-DH，
/ / AC = BD. (6 points)
AB is the chord of circle O. C, D are on AB, and AC = CD = dB. The extension line of OC and OD intersects circle O at E.F The results showed that the angle cod > angle AOC
Connect EF, easy to know EF parallel ab
The extension Fe and OA intersect at point G
According to the equal proportion theorem, AE = EF
Connect AE in triangle AEG
Gae-180 angle
GEA=180-AEF
AEF = OEA + OEF = OAE + OEF > OAE
Therefore, the angle gae > gea
Therefore, Ge > AE
EF>AE
Angle cod > angle AOC
In the circle O, AB is the diameter, AC is the chord, point D is on the chord AC, and OD = 5, ∠ ADO = 2 ∠ a = 60 ° and the length of CD is
Connect BC, because C is a point on the circumference, so ∠ ACB = 90 degrees, because ∠ ADO = 60, ∠ a = 30, so ∠ AOD = 90, so triangle ADO is similar to triangle ABC, because do = 5, Ao = 5, root sign 3, ab = 10 root sign 3, ad = 10, AD / AB = AO / AC, AC = 15, so CD = ac-ad = 5
As shown in the figure, OA is perpendicular to OC, ob is perpendicular to OD, and angle AOD = 2 angle BOC. Calculate the size of angle BOC
Because angle BOD equals angle AOD plus angle AOB equals 90 degrees. Angle AOC equals angle AOB and angle cob equals 90 degrees. And because angle AOD equals to angle BOC, it is very simple
As shown in the figure, AB is a chord of ⊙ o, OD ⊥ AB is at point D and point E is on ⊙ O. (2) if OC = 3, OA = 5, find the length of ab (2) If OC = 3, OA = 5, find the length of ab. (3) if P is a moving point on ⊙ o (which does not coincide with a and b), try to determine the quantitative relationship between ∠ APB and ∠ AOD? Please prove your conclusion
OC = 3?, without C, is it d? 2, ∵ AB is a chord of ⊙ o, and OD ⊥ AB is a chord of ⊙ o, and OD ⊥ AB is a chord of ⊙ o, and OD ⊥ AB is a chord of ⊙ o ⊥ AB at point dad = BD (any chord on the circle is vertically bisected by a vertical line passing through the center of the circle, which can be proved by congruent ⊥) ∵ OC = 3, OA = 5ad ∧ - od
As shown in the figure, the "O" and "O" intersect at two points a and B, point O is on "O", and the chord OC of O intersects AB at point D. It is proved that OA 2 = OC × OD
Proof: in "O"
∵ AB is the chord and OE is the radius
ν arc Ao = arc Bo
∵∠OAB=∠OCB,∠OCB=∠ACO
∴∠OAB=∠ACO
∵ OC is the diameter, OC ⊥ ab
∴∠OAC=90°,∠ODA=90°
∴△OAD∽△OCA
/ / OA ratio OC = od ratio OA
That is, OA 2 = OC × OD
A. B, C and D are on the same straight line, ab = CD, CE ⊥ ad, BF ⊥ ad, respectively. The vertical feet are C, D, connecting EF and crossing ad to g, AE ⊥ DF （1） Try to explain EF bisector BC (2) If △ BFD is translated along ad direction, other conditions remain unchanged, is the above conclusion still valid? (make do with it. I can't insert the picture, and I can't find it.) use ∵
（1）∵CE⊥AD BF⊥AD
Ψ ace = ∠ DBF = 90 ° (vertical definition)
And ∵ AB = CD
∴AB+BC=CD+BC
∴AC=BD
In RT △ ace and RT △ BDF
AC=BD
AE=BF
∴△ACE≌△BDF（HL）
∴BF=CE
In △ ECG and △ BFG
∠ACE=∠BDF
∠BGF=∠EGC
CE=BF
∴△ECG≌△BFG
∴BG=CG
ν EF bisection BC
(2) Similarly, it can be concluded that it is tenable
In the parallelogram ABCD, ad = 2Ab, extend AB to F, make BF = AB, extend Ba to e, make AE = AB, connect CE and DF to cross ad, BC to g, H CE ⊥ DF respectively Come on, write out the proof. Thank you
In triangle EBC and triangle EAG, because Ag / / BC, and ab = AE, 2ag = BC = ad, that is, G is the midpoint of AD; similarly, in triangle ADF and triangle BHF, because BH / / AD, and ab = BF, 2bh = ad = BC, so h is the midpoint of BC connecting GH, because GD = CD = ch = GH, so gdch is a diamond, so diagonals are mutually
Given the intersection point O, AB / / CD, OA = OD, AE = DF, prove be / / CF
How to solve it without drawing? Do you draw freely?
AB = CD, AE perpendicular to e, DF perpendicular to BC to F, ad crossed to o, OA = OD
The results and detailed process have been thought out in the exercise book. Give wealth first, then send the answer
RELATED INFORMATIONS
- 1. 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
- 2. 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
- 3. 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
- 4. 0
- 5. 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 °
- 6. In the hexagon ABCDEF, ∠ a = ∠ D, ∠ B = ∠ e, ∠ C = ∠ F, try to guess the position relationship between AF and CD, and explain the relationship
- 7. 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
- 8. 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．
- 9. 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 |
- 10. 0
- 11. 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
- 12. 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
- 13. In the triangle ABC, we know ∠ a = 30 ° and ∠ CBD = 90 ° to find the degree of ∠ BCE
- 14. 0
- 15. 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
- 16. 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
- 17. 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
- 18. 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
- 19. 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
- 20. 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