![](data:image/svg+xml;utf8,<svg xmlns="http://www.w3.org/2000/svg" xmlns:xlink="http://www.w3.org/1999/xlink" viewBox="0 0 72 71" width="72"></svg>)
As shown in FIG. 12, the diameter of circle O1 AB and chord CD are compared with point E. we know that AE = 6cm, EB = 2cm, ∠ CEA = 30 degrees. Find the length of CD
∵AE=6CM,EB=2CM
∴AB=AE+EB=8CM,OE=2CM
Make of perpendicular to CD, intersect CD and F
In RT △ ofe, ∠ CEA = 30 °
OF=（1/2）*OE=1CM
In RT △ ofd, DF = root (OD? Of? 2) = (root 15) cm
CD = 2 * DF = (2 pieces of 15) cm
As shown in the figure, in ⊙ o, AB is the diameter, chord CD ⊥ AB, e is a point on arc BC. If ⊙ CEA = 28 °, then ﹤ bad=______ .
∵ AB is the diameter,
∴∠ADB=90°，
∵ chord CD ⊥ AB,
Qi
AC=
AD，
∴∠CEA=∠B=28°，
∴∠BAD=90°-∠B=62°．
So the answer is: 62 degrees
AB is the diameter of ⊙ o, the chord CD ⊥ AB, e is a point on arc BC. If ᚜ CEA = 28 °, then ∠ CDB =?
∵ ab ⊥ CD, AB is the diameter
ν arc CD = arc ad
∵∠CAE=28°
∴∠ABD=28°
∵AB⊥CD
∴∠CDB=90°-28°=62°
As shown in the figure, AB is the diameter of circle O, BC is the chord of circle O, OD ⊥ CB is at e, intersection BC is at point D, and CD is connected, and the angle CDB = A and angle ABC = B are set. Try to find out the value between a and B And give a proof
Connecting ad
∠CDB=∠CDA+∠ADB
The circumference angle of the diameter is 90 degrees
So ∠ ADB = 90 °
The angle of the circle opposite the same arc is equal
∠CDB=∠ABC
∠CDB=90°+∠ABC
That is, a = 90 ° + B
It is known that AB is the diameter of ⊙ o, and the chords CD ⊥ AB, e are At a point on AC, the extension lines of AE and DC intersect at point F. it is proved that: ∠ AED = ∠ CEF
Proof: connect ad, as shown in the figure,
∵CD⊥AB，
/ / arc AC = arc ad,
∴∠ADC=∠AED，
∵∠CEF=∠ADC，
∴∠AED=∠CEF．
AB is the diameter of circle O, chord CD is vertical, AB, e is a point on arc AC, AE, CD extension line intersect with point F, it is proved that angle AED = angle CEF
Connecting EB
Because AB is the diameter and CD ⊥ ab
So arc BC and arc BD are equal (vertical diameter theorem)
So ∠ CEB = ∠ bed
Because AB is the diameter
So ∠ AEB = ∠ Feb = 90 degrees
Because ∠ AED = 90 - ∠ bed
∠CEF=90-∠CEB
So: ∠ AED = ∠ CEF
As shown in the figure, if the diameter ab of ⊙ o intersects with the chord CD at point E, AE = 5, be = 1, CD = 4, then ∠ AED=________ . Weihai City 2011 high school entrance examination mathematics question 15 picture can not upload ah.
The radius of the circle is 3 because of is perpendicular to CD. According to the vertical diameter theorem, DF = half of CD = 2 times the root 2. In the right triangle DOF, according to the Pythagorean theorem, we get of = 1
Given that the chords AB and CD of a circle intersect at point E, the degree of AC arc is 80 degrees, the degree of BD arc is 60 degrees, and the angle AEC is equal to how many degrees? The step diagram should be drawn by yourself One of the two solutions is 70 degrees, like the other is 170 degrees
Let the intersection e be in the circle
Connecting ad
∠AEC=∠ADC+∠BAD
Point E is outside the circle
70 degrees, 10 degrees and 110 degrees. I don't know if it's missing
The picture will be sent to you later
The chord AB and CD of circle O intersect at point E. arc AC = 60 degree arc BD = 40 degree angle AEC degree is calculated
Let AO and CD intersect at F do and ab intersect at g. in triangle CFO, angle FCO + angle COF + angle OFC = 180 degrees in triangle AFE, angle FAE + angle FEA + angle AFE = 180 degrees, because angle OFC = angle AFE and angle COF = 60 degrees, angle FAE + angle f
Given that the chord AD and CB of circle O intersect with point E, the degree of arc AC is 60, and the degree of arc BD is 100
Arc AC + arc BD = 160 degrees, so arc AB + arc CD = 360-160 = 200 degrees
The ∠ ACB and ∠ CAD are the circumferential angles of arc BC and arc ad respectively
So ∠ ACB + ∠ CAD = 100 degrees
In △ ace, ∠ AEC = 180 - (∠ ACB + ∠ CAD) = 80
RELATED INFORMATIONS
- 1. What is the relationship between the degree of angle AEC and the degree of arc AC and arc BD? If e is outside the circle, what is the relationship between the degree of angle AEC and the degree of arc AC and arc BD?
- 2. In the triangle ABC, ad is the center line on BC side, e is a point on AC, be and ad intersect F, if AE = EF, it is proved that BF = AC Here I know to extend ad, but I don't know if you can make CG through point C, hand over ad to g, make CG = DC, I don't know if this is OK. Ask!
- 3. As shown in the figure, in the triangle ABC, ab = AC, BF = CD, BD = CE, given the angle a = 40, calculate the degree of the angle EDF
- 4. At ⊙ o, C is the center of ACB, CD is the diameter, chord AB intersects CD at point P, PE is perpendicular to point E, if BC = 10, CE ratio EB = 3:2, AB is found
- 5. As shown in FIG. 11, AB is the diameter of circle O, passing through point o as parallel line of chord BC, tangent AP crossing point a, connecting point P and connecting AC. (1) verification: △ ABC ~ △ Poa (2) OB = 2, Op = 7 / 2, find the length of BC
- 6. As shown in the figure, in ⊙ o with diameter AB = 12, chord CD ⊥ AB is at m, and M is the midpoint of radius ob, then the length of chord CD is () A. 3 B. 3 Three C. 6 D. 6 Three
- 7. If point P is the point on the chord AB, connect OP, pass P as PC ⊥ OP, and PC intersect ⊙ o at point C. If AP = 4, Pb = 2, then the length of PC is () A. Two B. 2 C. 2 Two D. 3
- 8. The edge ab of the square ABCD with side length of 4 is the diameter of circle O, CF is the tangent line of circle O, e is the tangent point, f is the chord of circle O on AD 1 find the area of CDF 2 find the length of line segment be
- 9. As shown in the figure, AB and CD are the two chords of circle o
- 10. Given that the radius of circle O is 6 and the length of chord AB is 6 times the root sign 3, then the degree of the center angle of chord AB is?
- 11. As shown in the figure, AB is a chord of ⊙ o, OD ⊥ AB, perpendicular foot is C, intersect ⊙ o at point D and point E on ⊙ o (1) If ∠ AOD = 52 °, calculate the degree of ∠ DEB; (2) If OC = 3, OA = 5, find the length of ab
- 12. As shown in the figure, △ ABC, e is the heart of △ ABC, the bisector of ∠ A and the circumscribed circle of △ ABC intersect at point D. It is proved that de = dB
- 13. As shown in the figure, in the acute triangle ABC, AB > AC, ad is perpendicular to D, and graph o with AD as diameter intersects AB, AC with E and f respectively
- 14. In the triangle ABC, ab = 15, AC = 13, height ad = 12, find the circumference of triangle ABC
- 15. Take the right angle side of the RT triangle ABC as the diameter, make a semicircle o, intersect the oblique side with D, OE parallel with AC, and intersect AB with e. it is proved that De is the tangent of circle o
- 16. As shown in the figure, it is known that the triangle ABC is inscribed in the circle O, the point D is on the extension line of OC, SINB = 1 / 2, angle d = 30 ° (1) prove that ad is tangent. (2) if AC = 6, find the length of AD
- 17. The area of the inscribed triangle of radius 1 is 0.25. Find the product of the three sides of the triangle
- 18. In the isosceles triangle ABC, ∠ a = 120 ° AB = AC = 2cm, the minimum radius of circular paper covering △ ABC is calculated
- 19. As shown in the figure, in △ ABC, ∠ ABC = 2 ∠ C, ad is the height on BC edge, extend AB to point E, make be = BD, pass through point D, e lead straight line to AC at point F, then AF = FC, why?
- 20. Draw a circle with the vertex a of the square ABCD as the center and the side length as the radius. The area of the square is known to be 24 square centimeters. Calculate the area of the shadow part