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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 °
BD bisection ∠ ABC, CBD = 12 ∠ ABC, ? CD bisection △ ABC ? DCE = 12 ∠ ace = 12 (﹤ a + ∠ ABC) = 12 ∠ a + 12 ∠ ABC. In △ BCD, by the external angle property of triangle, ? DCE = ∠ CBD + ∠ d = 12 ∠ ABC + ∠ D, ? 12 ∠ a + 12 ∠ ABC = 12 ﹣ D, ﹤ d = 12 ∠ B
As shown in the figure, in △ ABC, the bisector of ∠ a = 90 ° intersects AB with D. if ∠ DCB = 2 ∠ B, calculate the degree of ∠ ADC
Let ∠ B = X,
∵∠DCB=2∠B，
∴∠DCB=2x，
The bisector of ∵ C intersects AB with D,
∴∠ACD=∠DCB=2x，
∵ ADC is the external angle of BCD,
∴∠ADC=∠B+∠DCB=3x，
In △ ACD,
∵∠A+∠ACD+∠ADC=180°，
ν 90 ° + 2x + 3x = 180 ° and x = 18 °,
∴∠ADC=3x=3×18°=54°．
If the point O is inside the triangle ABC and there is 4 vector OA + vector ob + vector OC = 0, then the ratio of the area of triangle ABC to the area of triangle OBC is 0
Let the midpoint of BC be d,
Vector ob + vector OC = 2 vector OD
∵ 4 vector OA+ vector OB+ vector OC= vector 0
ν 4 vector OA + 2 vector od = vector 0
Vector od = - 2 vector OA
So | a, O, D are collinear
|AD|=3/2|OD|
The area ratio of triangle ABC to triangle OBC is 3 / 2
Given that the point O is inside △ ABC, and vector OA + 2 vector ob + vector OC = 0, calculate the area ratio of △ ABC to △ AOC
I will not write the specific answer, I think you should be able to understand
If there is a point O in the triangle ABC and the vector OA · ob = ob · OC = OC · OA, is the point o the center of gravity, the outer center, the inner center or the perpendicular center of the triangle?
The above explanations are far fetched, or obscure
The correct explanation is:
From OA · ob = ob · OC
OA·OB-OC·OB=0
(OA-OC)·OB=0
Ca · ob = 0, that is, ob is perpendicular to the AC edge
Similarly, from ob · OC = OC · OA, OC is perpendicular to ab edge
From OA · ob = OC · OA, OA is perpendicular to BC edge
Obviously, the point O is perpendicular to the triangle
Given that point O is the center of gravity of triangle ABC, find vector OA + vector ob + vector OC =?
Point O is the center of gravity of triangle ABC = = > center line ad, be, CF, and vector Ao = 2 vector OD, vector Bo = 2 vector OE, vector co = 2 vector of
The interior point O of the triangle ABC satisfies the following conditions: a vector OA + B vector ob + C vector OC = 0 vector
Let the radius of the inscribed circle of △ ABC be r, then s △ BOC = (1 / 2) * a * r = (1 / 2) * r = (1 / 2)) * (1 / 2) * |ob ||||||||||||||||||||||||||||||||||||||||||||||||/ R) * sin ∠ Aoba * OA + b * ob + c * OC = (| ob | * | OC | / R) * sin ∠ BOC * OA + (| OC | o
As shown in the figure, O is a point in the triangle ABC. Try to explain that OA + ob + OC > half (AB + BC + Ca)
For a triangle, the sum of the two sides is greater than the third
OA+OB>AB; (1)
OA+OC>AC; (2)
OB+OC>BC; (3)
Then (1) + (2) + (3), 2 (OA + ob + OC) > AB + AC + BC
OA + ob + OC > 1 / 2 * (AB + AC + BC)
Given that O is any point in the triangle ABC, we try to explain: (1) one half of (AB + AC + BC) OA + ob + OC
(1)OA+OB>AB
OB+OC >BC
OA+OC>AC
2(OA+OB+OC)>AB+AC+BC
OA+OB+OC>1/2(AB+AC+BC)
It is known that the point O is inside the triangle ABC, connecting OA, ob, OC, indicating that, half (AB + AC + BC)
According to the fact that the sum of the two sides is greater than the third side, for the triangle OAB, OBC, OAC, there are: OA + ob > AB; OA + OC > AC, OB + OC > BC; therefore, OA + ob + OA + OC + ob + OC > AB + AC + BC; therefore, half (AB + AC + BC) < OA + ob + OC; next prove that OA + ob + OC < AB + AC + BC
RELATED INFORMATIONS
- 1. 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
- 2. In the triangle ABC, BC = 2, Sina = 2 / 3, root 2, then what is the maximum product of AB vector and AC vector
- 3. In △ ABC, Tanc = 3, radical 7 (1) find COSC (2) if vector CB · vector CA = 5 / 2 and a + B = 9 find C Process
- 4. In the triangle ABC, a = 60 degrees, AC = 16, area s = 220, root number 3, then BC = what?
- 5. An obtuse triangle ABC, ∠ B is an obtuse angle, where AB = 10, BC = 9, AC = 17, find the height on BC side
- 6. In RT △ ABC, ∠ C = 90 °, ab = 8, AC = 4 times the root sign 3, find the degree of ∠ a, ∠ B
- 7. In the triangle ABC, AB is equal to AC, angle a is equal to 20 degrees, and point D is on AB, so that ad is equal to BC, and the degree of angle BDC is obtained Note: this problem only isosceles triangle and congruent triangle to prove
- 8. In the triangle ABC, if a + C = 2B, C = 2, a = radical 3 + 1, find the value of B as in the title
- 9. As shown in the figure, △ ABC, D is on AC, ad: DC = 1:2, e is the midpoint of BD, and the extension line of AE intersects BC at F, Verification: BF: FC = 1:3
- 10. It is known that: as shown in the figure, △ ABC (AB ≠ AC), D and E are on BC, and de = EC, DF ‖ Ba is made through D, and AE is intersected with point F, DF = AC. it is proved that AE bisection ∠ BAC
- 11. 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
- 12. 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°
- 13. 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
- 14. 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
- 15. In the triangle ABC, if AC = radical 2, BC = radical 7, ab = 3, then cosa =?
- 16. 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______ .
- 17. 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
- 18. 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
- 19. 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?
- 20. 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