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Given that AB and CD are two diameters of circle O, chord CE / / AB, angle COE = 50 degrees, then the acute angle DOB =?
Solution: CO = OE = radius
The △ COE is an isosceles triangle
∴∠OCE=∠CEO
∠COE=50° ∴∠OCE=∠CEO=（180-50）/2=65°
AB‖CE
∴∠OCD=∠DOB
∴∠DOB=65°
A. B is the intersection point of circle O1 and circle O2. AC is the diameter of circle O1. The extension line of Ca and CB intersects circle O2 at D, e. AC = 12. Be = 30. BC = ad Find the length of de
According to the secant theorem, ca * CD = CB * ce, AC = 12, be = 30, let BC = ad = x, then 12 (12 + x) = x (x + 30), ν x ^ 2 + 18x-144 = 0, x > 0, ν x = 6, BC = AC / 2, ∠ C = 60 °, CD = Ca + ad = 12 + 6 = 18, CE = CB + be = 6 + 30 = 36 = 2CD, ∠ d = 90 °, de = 18 √ 3
As shown in the figure, it is known that ⊙ O1 and ⊙ O2 intersect at points a and B, the lines passing through point a intersect two circles at points c and D, and the lines passing through point B intersect two circles at points E and f respectively, and ef ∥ CD. Verification: CE = DF
Proof: connect ab,
∵CD∥EF，
Qi
CE＝
AB，
∴CE=AB，
Similarly, ab = DF,
∴CE=DF．
Circle O1 and circle O2 intersect at a and B, the straight line passing through a intersects two circles at C and D, M is the midpoint of CD, and BM intersects circle E and F (1) Verification: CE parallel DF (2) Confirmation: me = MF 2. Circle P and circle O intersect at two points a and B. circle P passes through the center O and C is any point on the superior arc AB of circle P (not coincident with a and b), connecting AB, AC and OC (1) Point out the angle equal to the angle ACO (2) when C is in the circle P, the straight line AC is tangent to the circle O, reason (3) when the angle ACB = 60 degrees, what is the relationship between the radius of the two circles
prove:
1. Connect ab
Because C = B, d = B
(equal to the circle angle of the arc)
So, C = D
So CE / / DF
2、
Because m is the midpoint of the CD
So cm = DM
And because ∠ C = ∠ D, ∠ CME = ∠ DMF
So △ CEM ≌ △ DFM (ASA)
So me = MF
⊙ O1 and ⊙ O2 intersect at points a and B. the straight line passing through a intersects two circles at points c and D, and the line passing through B intersects two circles at points E and F. moreover, CD ∥ EF Confirmation: CE = DF Fast, fast, fast, online, etc
Proof: connect ab
∵ ABFD is a quadrilateral inscribed in a circle
∠ABF+∠D=180°
ABEC is a quadrilateral inscribed in a circle
∴∠C=∠ABF
∴∠C+∠D=180°
∴AE‖DF
∵CD‖EF
The aefd is a parallelogram
∴CE =DF
It is known that: O 1, O 2, intersect at two points a and B, the straight line passing through point a is perpendicular to AB and intersects two circles at point C, D. It is better to have a graph to prove CD = 2o1o2
It is proved that o1e is perpendicular to CD to e, o2f is perpendicular to CD to F,
Then AE = AC / 2, AF = ad / 2 (vertical diameter theorem)
So EF = 1 / 2CD,
Because circle O1 and circle O2 intersect at two points a and B,
So O1O2 divides AB vertically,
Because CD is perpendicular to AB,
So O1O2 / / CD,
Because o1e is perpendicular to CD, o2f is perpendicular to CD,
So o1e / / o2f,
Therefore, EF = O1O2 (the parallel segments between two parallel lines are equal),
So O1O2 = 1 / 2CD,
That is, CD = 2o1o2
It's easy to draw this picture, but my computer can't send it,
Circle O1 and circle O2 intersect at a and B, the straight line passing through point a intersects two circles at point C and D, and the line passing through point B intersects two circles at e and f respectively, and CD / / EF Confirmation: CE = DF
GH ∥ CD is made over the center of the circle O1, and ∠ co1e = ∠ ao1b is proved by GH,
∴CE=AB
Similarly, ab = DF
CE=DF
As shown in the figure, it is known that ⊙ O1 and ⊙ O2 intersect at points a and B, lines CD and EF cross point B, intersect ⊙ O1 at points c and E, and ⊙ O2 at points D and F (1) Verification: △ ACD ∽ AEF; (2) If ab ⊥ CD and the length of AF, AE and EF in ⊥ AEF are 3,4,5 respectively, it is proved that AC is tangent of ⊙ O2
It is proved that: (1) in ⊙ O1, ∵ C = ∵ e, ∵ d = ∵ F, ∵ ACD ∵ AEF; (2) ∵ ab ⊥ CD, i.e. ∵ abd = 90 °, ad is the diameter of ⊙ O2, ∵ in ? AEF, af2 + AE2 = 32 + 42 = 52 = ef2, ∵ EAF = 90 ° is obtained from (1) M CAD = ? EAF
As shown in the figure, it is known that circle O1 and circle O2 intersect at points a and B, lines CD and EF pass through point B, circle O1 intersects point C, and circle O2 intersects points D and F If ab ⊥ CD, and in △ AEF, the lengths of AF, AE and EF are 3,4,5, respectively Verification: AC is tangent of circle o
It is proved that ∵ ab ⊥ CD is ∵ abd = 90 °,
⊙ ad is the diameter of ⊙ O2,
In △ AEF, af2 + AE2 = 32 + 42 = 52 = ef2,
∴∠EAF=90°,
⊙ in ⊙ O1, ᙽ C = ᙽ E,
∵∠D=∠F,
∴△ACD∽△AEF
∴∠CAD=∠EAF=90°,
∴AC⊥AD,
And ∵ ad is the diameter of ⊙ O2,
/ / AC is the tangent of circle o
As shown in the following figure, circle O1 and circle O2 intersect at two points a and B, the straight line CD passing through point a intersects circle O1 at C, intersects with circle O2 at D, and the line EF passing through point B intersects circle O1 As shown in the following figure, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects circle O1 at C and circle O2 at D. the line EF passing through point B intersects circle O1 at e and circle 02 at F. it is proved that CE ∥ DF, me = MF
prove:
(1) From the diagonal complementarity of the inscribed quadrilateral of a circle, it can be seen that:
∠ CEB + ∠ cab = 180 degrees, ∠ BFD + ∠ bad = 180 degrees
That is, ∠ CEB + ∠ cab + ∠ BFD + ∠ bad = 360 degrees
It is easy to know that ∠ cab + ∠ bad = 180 degrees
Therefore, ∠ CEB + ∠ BFD = 360 degrees - 180 degrees = 180 degrees
The CE ∥ DF can be obtained by complementing the internal staggered angles
(2) Where is the m point
Digression: if CE ∥ DF is not proved in the examination, but it can still be used as the known condition to prove me = MF, the same score can be obtained
RELATED INFORMATIONS
- 1. It is known that, as shown in the following four graphs, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects with circle O1 at C, intersects with circle O2 at D, and straight line ef? 2 passing through point B As shown in the following figure, circle O1 and circle O2 intersect at two points a and B, the line CD passing through point a intersects with circle O1 at C, and with circle O2 at D. the line EF passing through point B intersects with circle O1 at e and circle 02 at F, so as to verify CE ∥ DF Picture in space photo album
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