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Known: as shown in the figure, in the right triangle ABC, the angle c = 90 degrees, the point O is the center of the circle, the length of OA is the radius of the circle, and AC, AB intersect with point D.E, and the angle CBD = angle a.bd is the tangent line of circle O. find: if ad: Ao = 8:5, BC = 2, find the length of BD
Oh is perpendicular to ad, because DB is tangent, so ∠ BDO = 90 degrees, so ∠ OH + ∠ BDC = 90, because ∠ CBD + ∠ BDC = 90, so ∠ ODH = ∠ CBD, because ∠ C = 90, so triangle BDC is similar to triangle DOH, so BC: DB = DH: do / x0d because OA = do, oh is the height of AD, ah = DH, so DH: do = 4:5, so BC: DB = 4:5, so BD = 5 / 4bc
As shown in the figure, the known point O is a point on the slope ab of RT △ ABC. The circle with o as the center and OA as the radius is tangent to point D and intersecting with ab at point E (1) Try to judge whether AD bisects ∠ BAC? And explain the reason (2) If BD = 3bE, CD = 3, find the radius of ⊙ o
(1) Judgment: ad bisection ∠ BAC
prove:
Method 1: connect OD;
D ⊙ BC ⊙,
∴OD⊥BC，
And △ ABC is RT △ and ∠ C = 90 °,
∴AC⊥BC，
∴OD∥AC，
∴∠1=∠2；
And ∵ OA = OD,
∴∠3=∠2，
∴∠1=∠3．
Method 2: connect ed;
∵ AE is ⊙ o diameter,
∴∠ADE=90°，
∴∠3+∠AED=90°；
And ∵ C = 90 °,
∴∠1+∠ADC=90°，
And ? AED = ∠ ADC,
∴∠1=∠3．
Method 3: connect EF and DF;
∵ AE is ⊙ o diameter,
∴∠AFE=90°，
And ∵ ace = 90 °,
∴∠AFE=∠ACB，
∴EF∥BC，
∴∠4=∠5；
And ? 3 = ∠ 4, 1 = ∠ 5,
∴∠1=∠3．
（2）
Solution 1: let be = x, then BD = 3bE = 3x,
According to the cutting line theorem, BD2 = be × ba,
AB = 9x, OA = OE = 4x;
And ∵ OD ∫ AC,
∴OB
OA＝BD
CD, i.e. 5x
4x＝3x
3，
∴x=5
4，
The radius of ⊙ o is 5
Solution 2:
As shown in the figure, og ⊥ AC, AC ⊥ BC, OD ⊥ BC,
Then the odcg of quadrilateral is rectangle
∴OG=CD=3，OG∥BC；
Og ∥ BC,
∴OG
BC＝OA
AB，
∴3
3x+3＝4x
9x，
∴x=5
4, x = 0, (omitted)
The radius of ⊙ o is 5
Note: this solution is based on AB = 9x, OA = OE = 4x
As shown in the figure, the radius of circle O is 1, and the isosceles right triangle ABC
The problem is not complete. And there is no picture
In the triangle ABC, the angle BAC = 90 degrees AB = AC = 2 and root sign 2. Take a as the circle point with radius 1 and move on the edge of BC (not coincide with a and b). Take o as radius to make circle O. when circle O is tangent to circle a, what is Bo equal to?
(1) In ∵ ABC, ∵ BAC = 90 °, ab = AC = 22,
ν Δ ABC is an isosceles right triangle,
∴BC=4,
⊙ A and BC are tangent to point D,
∴AD=r,AD⊥BC,
The ad is the center line on the BC side,
∴r=AD= 12BC=2,
(2) (1) make ad ⊥ BC at point D,
∵ △ ABC is an isosceles right triangle, BC = 4,
The ad is the center line on the BC side,
∴AD= 12BC=2,
∴S△AOC= 12OC•AD,
∵ Bo = x, the area of ∵ AOC is y,
∴y=4-x（0＜x＜4）,
② Make OE ⊥ AB through point O and cross AB to E,
⊙ the radius of a is 1, OB = X,
When two circles are circumscribed,
∴OA=1+x,
∵ △ ABC is an isosceles right triangle,
∴∠B=45°,
∴BE=OE= 22x,
oe2 ﹤ oe2 ﹣ 2 ﹣ in AE2 ﹤ oe2 ﹣ 2 ﹣ AE ﹣ 2 ﹣ e ﹤ 2 ﹤ a ﹤ o,
∴（1+x）2=（2 2- 22x）2+（ 22x）2,
∴x= 76,
∵ △ AOC area = y = 4-x,
The area of △ AOC = 176;
When two circles are inscribed,
∴OA=x-1,
∵AO2=AE2+OE2=（AB-BE）2+OE2,
∴（x-1）2=（2 2- 22x）2+（ 22x）2,
∴x= 72,
The area of △ AOC = y = 4-x = 4 - 72 = 12,
The area of AOC is 176 or 12
In the triangle ABC, ab = radical three, AC = 2. If O is a point inside △ ABC and the vector OA + ob + OC = 0, then the vector AO * BC =?
AB = OB -OA ,AC = OC -OA ,∴ AB + AC = OB + OC -2 OA =-3 OA ,
∴ OA =-AB + AC 3 ,
If we make a parallelogram abdc, then o is at the trisection point of AD and is close to a,
Therefore, the AOB area is 1 / 3 of half of the abdc area, i.e. ABC
1.3% of the area,
∵ ab ᙽ AC = 2,3, | | ab | AC | = 4, ᙽ △ ABC area is 1,
The area of △ oba is 13
O is the outer center of the triangle ABC, e is a point in the triangle, and the vector OE = vector OA + vector ob + vector OC RT proves that the vector AE is perpendicular to the vector BC (or perpendicular to the center of E)
Vector AE = vector OE - vector OA = vector ob + vector OC (derived from known conditions)
Vector BC = vector oc - vector ob
Then we have vector AE * vector BC = OC squared - ob squared = 0 (o is the outer center OC = OB)
AE vertical BC
As shown in the figure, in △ ABC, ∠ a = 50 ° o is a point in △ ABC, and ∠ ABO = 20 ° and ∠ ACO = 30 ° are used to calculate the degree of ∠ BOC
Extend Bo to AC to E,
∵∠A=50°，∠ABO=20°，
∴∠1=50°+20°=70°，
∵∠ACO=30°，
∴∠BOC=∠1+∠ACO=70°+30°=100°
It is known that in the figure △ ABC, AC = BC, ∠ ACB = 80 ° o is a point in △ ABC, ∠ OAB = 10 ° and ∠ oba = 30 °, then ∠ ACO= If you have an auxiliary line, you 'd better have a picture
It is easy to get △ CAQ = 60 ° - 50 ° = 10 ° = ∠ OAB; ∠ QCB = 80 ° - 60 ° = 20 °; CQ = CA = CB, so ∠ CBQ = 80 °, ABQ = ∠ CBQ - ∠ CBA = 80 ° - 50 ° = 30 °, so ∠ ABQ = ∠ oba, so △ OAB ≌ △ qAB, so Ao = AQ = AC and ∠
In the triangle ABC, AB is the bottom, AC is equal to BC, angle ACB is 80 degrees, O is a point in the triangle, OAB is 10 degrees, and angle ABO is 30 degrees. What is the angle ACO?
In the triangle ABC, AC = CB, angle ACB = 80 degrees, point O is a point in the triangle, and the angle OAB = 10 degrees, angle ABO = 20 degrees, the degree of angle ACO is calculated
AC = BC = 5, angle ACB = 80 degrees, O is a point in ABC, angle OAB = 60 degrees, angle oba = 30 degrees, then AB length kuai!
It's impossible. The angle cab = 50 degrees. How can the angle OAB be 60 degrees?
RELATED INFORMATIONS
- 1. Point O is a point in △ ABC, ∠ ABO = 25 degrees, ∠ ACO = 35 degrees, and the degree of ∠ BOC is 5 degrees less than 2 times of ∠ a In △ ABC, the top angle is ∠ a, the left is ∠ ABC, and the right is ∠ ACB
- 2. In △ ABC, if the bisectors of ∠ a = 50 °, B and C intersect at point O, then the degree of ∠ BOC is () A. 65° B. 100° C. 115° D. 130°
- 3. As shown in the figure, in the RT triangle ABC, the angle B = 90 degrees. If the distance between the point O and the triangle three times is equal, what is the angle AOC? For detailed explanation
- 4. It is known that: as shown in the figure, in △ ABC, the bisector of ∠ B and the bisector of external angle of △ ABC intersect at point D, ∠ a = 90 °
- 5. It is known that the distance between point O and AB and AC is equal, and ob = OC (1) As shown in Figure 1, if point O is on edge BC, it is proved that ab = AC; (2) As shown in Figure 2, if point O is inside △ ABC, it is proved that ab = AC; (3) If point O is outside △ ABC, is ab = AC true? Please draw a picture to show it
- 6. In the triangle ABC, BC = 2, Sina = 2 / 3, root 2, then what is the maximum product of AB vector and AC vector
- 7. In △ ABC, Tanc = 3, radical 7 (1) find COSC (2) if vector CB · vector CA = 5 / 2 and a + B = 9 find C Process
- 8. In the triangle ABC, a = 60 degrees, AC = 16, area s = 220, root number 3, then BC = what?
- 9. An obtuse triangle ABC, ∠ B is an obtuse angle, where AB = 10, BC = 9, AC = 17, find the height on BC side
- 10. In RT △ ABC, ∠ C = 90 °, ab = 8, AC = 4 times the root sign 3, find the degree of ∠ a, ∠ B
- 11. In the triangle ABC, if AC = radical 2, BC = radical 7, ab = 3, then cosa =?
- 12. As shown in the figure, it is known that ∠ AOB = 30 ° and the point P is inside ∠ AOB, Op = 6. If there is a moving point m on OA and a moving point n on ob, then the minimum perimeter of △ PMN is______ .
- 13. As shown in the figure, the side of the cone is cut along OA, and the fan-shaped OAB is expanded into a plane figure (1) What is the relationship between the length of the sector arc AB and the length of the circle at the bottom of the cone? What is the position relationship between point a and point B on the side of the cone? (2) If the angle ∠ AOB = 90 °, what is the relationship between the radius r of the cone bottom circle and the radius r of fan-shaped OAB (i.e. OA or OB)? (3) If point a moves on the side of the cone and then returns to its original position, how should the shortest distance of point a be designed? If R 2 = 0.5, ∠ AOB = 90 °, find the shortest distance of point a
- 14. The right angle vertex P of the triangle ruler slides on the ray OM, and the two right angle sides intersect OA and ob at point CD respectively Conjecture the quantitative relationship between PC and PD
- 15. As shown in the figure, OA, OB and OC are the radii of circle O. if ∠ AOB = 2 ∠ BOC, please judge whether ∠ ACB = 2 ∠ BAC is tenable. Why?
- 16. As shown in the figure, the ray OA, ob, OC, OD have common endpoint o, and ∠ AOB = 90 °, COD = 90 °, AOD = 5 4 ﹤ AOC. calculate the degree of ﹤ BOC
- 17. In the plane rectangular coordinate system, the image of function y = - 3 / 4x + 6 intersects X axis and Y axis at point a and B respectively, line BC and X axis intersect at C, and point C is the midpoint of line OA Find out whether there is a point m on the straight line AB so that △ BCM is an isosceles triangle. If there is, find out the coordinates of point M
- 18. Known: as shown in the figure, find a point P on the straight line Mn, so that the distance between the point P and the two sides of ∠ AOB is equal (the method is required to be written, the drawing trace shall be retained, and the conclusion shall be written)
- 19. As shown in the figure, ∠ AOB = 30 ° OC bisection ∠ AOB, P is a point on OC, PD ‖ OA intersects ob at point D, PE ⊥ OA at e, OD = 4cm, then PE ⊥ OA is at point E=______ .
- 20. The bottom surface of four pyramids P-ABCD is square, and the bottom surface of PB ⊥ ABCD proves that no matter how the height of four pyramids PB changes, the surface PAD and the surface PCD cannot be vertical