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If the tangent point of a circle is a tangent point of a circle, then the tangent point of the circle is abo Take yourself in the picture below
[analysis: if point E is different from point a and D, then ∠ e is ∠ AED] ? CB tangent circle O at point B ABC = 90 ? C = 40 ? bad = 50 ° connect BD ? AB is diameter ? ADB = 90 ° then ? abd = 90 ° - ｛ bad = 40 ° abd = 40 °
As shown in the figure, BD is the diameter of ⊙ o, points a and C are on ⊙ o, ab = AC = 2 3, the chord ad intersects BC at point E, and ad = 6 (1) Find the degree of ∠ ABC and the length of segment be; (2) It is proved that BF = Bo
(1) ∵ BD is the diameter,
∴∠BAD=90°，
∵AD=6，AB=2
3. According to Pythagorean theorem: BD=
AB2+AD2=4
3，
∴AB=1
2BD，
∴∠D=30°，
∴∠C=∠D=30°，
∵AB=AC，
∴∠ABC=∠C=30°，
∵∠BAD=∠BAE=90°，∠D=∠ABE=30°，
∴△ABE∽△ADB，
∴AB
AD=BE
BD，
∴2
Three
6=BE
Four
3，
∴BE=4．
(2) Proof:
Connect OA,
∵∠D=30°，
∴∠AOB=2∠D=60°，
∵OA=OB，
The △ AOB is an equilateral triangle,
∴AB=OB，∠OAB=∠ABO=60°，
∵ AF cut ⊙ o in a,
∴∠OAF=90°，
∴∠FAB=90°-60°=30°，
∴∠F=∠ABO-∠FAB=60°-30°=30°=∠FAB，
∴FB=AB，
∵AB=BO，
∴BF=BO．
It is known that AB is the diameter of circle O, CB ⊥ AB is in B, AC intersects circle O in D, de tangent circle O, circle D intersects BC and e. it is proved that de2 = 1 / 4CD * ca As shown in the figure, AB is the diameter of circle O, CB ⊥ AB is in B, AC intersects circle O in D, de cuts circle O, circle D intersects BC in E. verification: de2 = 1 / 4CD * Ca (please draw a picture yourself)
The diameter of the circle O is the diameter of the circle O, and the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O. the diameter of the circle O is the diameter of the circle O is the diameter of the connecting BD 888787878787878787? B 87878787878787878757cd * CA = BC ^ 2
As shown in the figure, CB cuts ⊙ o at point B, CA intersects ⊙ o at point D, and ab is the diameter of ⊙ o, and point E is A point on abd which is different from points a and D. if ∠ C = 40 °, then the degree of ∠ e is______ .
As shown in the figure: connect BD,
∵ AB is the diameter,
∴∠ADB=90°，
∵ BC cut ⊙ o at point B,
∴∠ABC=90°，
∵∠C=40°，
∴∠BAC=50°，
∴∠ABD=40°，
∴∠E=∠ABD=40°．
So the answer is: 40 degrees
As shown in Figure 5, AB is the diameter of circle O, point C is a point on Ba extension line, CD tangent circle O to point D, chord De is parallel to CB, q is a point on AB, CA = 1, CD = root sign 3oa Radius of circle o r the area of the shadow part in the graph
According to the meaning of the title, △ ODC is a right triangle,
So, OD ^ 2 + CD ^ 2 = OC ^ 2
Because od = R, OC = R + 1, CD = √ 3 × R
Therefore, R ^ 2 + (√ 3R) ^ 2 = (R + 1) ^ 2
R^2+3R^2=(R+1)^2
4R^2=(R+1)^2
(2R)^2=(R+1)^2
2R=R+1
R=1
Because there is no map, I don't know the shadow range,
AB is the diameter of circle O, ad is Xuan, DAB = 22.5 degrees. Extend AB to point C such that ∠ ACD = 45 degrees AB is the diameter of circle O, ad is Xuan, DAB = 22.5 degrees. Extend AB to point C so that ∠ ACD = 45 degrees. (1) prove that CD is tangent of circle O? (2) weak AB is the tangent of circle O. find the length of BC? In addition, find the meaning of sin and COS
Let o be the center of the circle, and connect OD (1) at △ AOD, from OA = OD, we can get ∠ DAB = ∠ ADO = 22.5 degrees, at △ cod, ∠ cod = ∠ DAB + ∠ ADO = 45 degrees, then: ∠ ODC = 180-45-45 = 90 degrees, so: od vertical DC, CD is tangent line of circle O! (2) at △ cod, OC = od / sin45 = 2 BC = oc-ob = 2-radical 2
It is known that: as shown in the figure, AB is the tangent line of ⊙ o, the tangent point is a, OB intersects ⊙ o at C and C is the midpoint of ob, and the chord CD passing through C makes
Connect OA, OD
∵∠DCA=45°
∴∠AOD=90°
Qi
The length of ad is 90 π· OA
180＝
Two
2 pi
∴OA=OD=
Two
∴AD=
OA2+OD2＝
4=2
∵ AB is ⊙ o tangent
∴OA⊥AB
C is the mid point of RT △ AOB beveled edge
∴AC=OC=OA=
2．
As shown in the figure, be bisection ∠ ABC, CE bisection ∠ ACD is known, and be is given to E. verification: AE bisection ∠ fac
It is proved that: as shown in the figure, the crossing point E is eg ⊥ BD, eh ⊥ Ba, EI ⊥ AC, and the vertical feet are g, h and I respectively,
∵ be bisection ∵ ABC, eg ⊥ BD, eh ⊥ Ba,
∴EH=EG．
∵ CE bisection ∵ ACD, eg ⊥ BD, EI ⊥ AC,
∴EI=EG，
Ψ EI = eh (equivalent substitution),
Ψ AE bisection ∠ fac (points with equal distance to both sides of the corner must be on the bisector of the angle)
As shown in the figure, the bisector of ab ∩ CD, ∩ BAC and ∩ ACD intersect at point E. verification: AE ⊥ CE
AB\CD \\\\\\BAC+ \\\\ACD=180 \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\-90 ° =90 ° is AE ⊥ ce
Known, as shown in the figure, ab ‖ CD, AE bisection ∠ BAC, CE bisection ∠ ACD, find the degree of ∠ E
∵ ab ∥ CD, AE bisection ∵ BAC, CE bisection ∠ ACD,
And ∠ BAC + ∠ DCA = 180 ° {CAE + ∠ ace = 1
2（∠BAC+∠DCA）=90°，
∠E=180°-（∠CAE+∠ACE）=90°，
∴∠E=90°．
RELATED INFORMATIONS
- 1. Known, as shown in the figure, ab ‖ CD, AE bisection ∠ BAC, CE bisection ∠ ACD, find the degree of ∠ E
- 2. As shown in the figure BD is the diameter of circle O, ab = AC, ad intersects BC at point E, AE = 2, ed = 4 The first question is to find the length of AB, the second question is to extend DB to f so that BF = Bo, connect FA, and prove that FA is tangent to circle o
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- 4. As shown in Figure AB is the diameter of circle O, ad and circle O are tangent to point a, de and circle O are tangent to point E, and point C is a point on the extension line of De, and CE = CB is connected to the extension of AE and AE
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