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In ABC, BD is the inner bisector of angle B, CE is the outer bisector of angle c, AF is perpendicular to BD, Ag is perpendicular to CE, FG and ABC?
What you may want to ask is the relationship between FG and △ ABC
FG‖BC；
It is proved that the extension of AF intersects BC with H, because BF is the bisection angle B of △ ABH vertical line, then f is the middle point of ah;
Make AE ‖ BC and hand it to CE in E;
Then ∠ ace = ∠ ECL = ∠ AEC, Ag ⊥ EC, then G is the midpoint of EC
FG is the median line of echelon aech, ∥ FG ∥ BC
DB and CE are bisectors of triangle ABC respectively. AF is made through a, BD is perpendicular to F, Ag is perpendicular to CE is perpendicular to g, and FG = 1 / 2 (AB + CB + AC)
Either under this condition, FG = 1 / 2 (AB + ac-bc); or dB and CE are bisectors of the exterior angles of ABC, FG = 1 / 2 (AB + BC + AC)
Here we prove the first conclusion
Lengthen AF and Ag to BC in M and N, respectively
BD bisects ∠ ABC (i.e. ∠ ABM), so BD is the bisector of △ ABM angle
AF ⊥ BD, so BD is also the height of am on the side of △ ABM, so △ ABM is an isosceles triangle
AB = BM, f is the midpoint of am
Similarly, CE is the bisector of △ ACN angle and the height of △ ACN side an, so △ ACN is an isosceles triangle
AC = CN, G is the midpoint of an
MN=BM+CN-BC=AB+AC-BC
F. G is the midpoint of AM and an respectively, so FG is the median line of △ amn
FG=MN/2=1/2（AB+AC-BC）
In △ ABC, where ∠ C = 90, CA = CB, e and F are respectively on AB and ∠ ECF = 45, can the line AE EF FB form a RT triangle
Yes,
Rotate the triangle BCF 45 degrees around C until ACD, AC and BC coincide and connect De,
Triangle BCF congruent triangle ACD, BF = ad,
It can be proved that the triangle DCE congruent triangle FCA, de = EF,
∠DAC+∠CAE=90
So can the segment AE EF FB form a RT triangle
Angle ACB = 51 degrees, AE perpendicular to AB, CM perpendicular to AB, BF perpendicular to AB, AE = BM, calculate angle ECF
In △ ABC, the angle ACB is 90 ° and CD is perpendicular to AB and D. take CD as radius and make circle C, tangent to AE at point E. BM ⊥ cm, that is BM / / AE, BM is tangent to circle C
It is known that a line L: x = 2p intersects a and B with a parabola y & # 178; = 2px (P > 0). It is proved that OA ⊥ ob
Substituting x = 2p into Y & # 178; = 2px leads to
y^2=4p^2,
Y = soil 2p,
∴A(2p,2p),B(2p,-2p),
The vector OA * ob = (2P) ^ 2 - (2P) ^ 2 = 0,
∴OA⊥OB.
Given that the line y = - x + A and the parabola y = x ^ 2 intersect at two points a and B, O is the origin, the scalar product of OA vector and ob vector is obtained
Given that the line y = - x + A and the parabola y = x ^ 2 intersect at two points a and B, O is the origin, the scalar product of OA vector and ob vector is obtained
Analysis: let a (x1, Y1), B (X2, Y2)
x^2=-x+a==>x^2+x-a=0
According to Weida's theorem: X1 + x2 = - 1, x1x2 = - A
Y1y2=(-x1+a)(-x2+a)=x1x2-a(x1+x2)+a^2=-a+a+a^2=a^2
The vector OA &; the vector ob = x1x2 + y1y2 = - A + A ^ 2 = a ^ 2-A
There is a parabola shaped bridge opening. The maximum height BM of the bridge opening from the water surface is 3 meters, and the span OA is 6 meters. The rectangular coordinate system is established with the straight line where OA is located as the x-axis and o as the origin (as shown in the figure)
(1) write the function analytic formula of parabola;
(2) a small boat has some rectangular planks which are 3 meters long, 2 meters wide and of uniform thickness. In order to make the boat pass through the bridge, how many meters can these planks be piled up at the highest level?
The graph is like this: the parabola of the function image opens downward, passes through (0,0) and (6,0), and the vertex coordinates are (3,3)
Let the equation y = ax ^ 2 + BX + C pass through the point (0,0) (6,0), and (3,3) be substituted into C = 00 = 36a + 6b3 = 9A + 3b to calculate a = - 1 / 3, B = 2, the analytic expression of image function y = - x ^ 2 / 3 + 2x (2) control the width 2, then the image can be vertically translated downward to get y = - x ^ 2 / 3 + 2x-c. let the solution of the equation be x1, X2, then | x1-x2 | = 2 (x1-x2) ^
The known parabola y = - x square + 2 (k-1) x + K + 2
It intersects with X axis at two points AB, and point a is on the negative half axis of X, and point B is on the positive half axis of X
Question: (1): find the value range of real number K
(2) Let the lengths of OA and ob be a and B respectively, and a: B = 1:5
y(0)=k+2>0
k>-2
Let a (- N, 0), B (5N, 0)
x1+x2=2(k-1)=4n
x1*x2=-(k+2)=-5n^2
K = (radical 6-1) / 2
Y = - x ^ 2 + (radical 6-1) x + (radical 6 + 3) / 2
As shown in the figure, the straight line y = 3x + 3 intersects the X axis at point a, and intersects the Y axis at point B. take AB as the right angle side, make isosceles RT △ ABC, ∠ BAC = 90 °, AC = AB, and the hyperbola y = KX passes through point C
① Find the analytic formula of hyperbola; 2. Point P is a point on the fourth quadrant hyperbola, connecting BP, point Q (x, y) is a moving point on line AB, QD ⊥ BP through Q, if QD = n, ask whether there is a point P such that y + n = 3? If it exists, find the BP analytic formula of straight line; if it does not exist, explain the reason
① From y = 3x + 3, a (- 1,0), B (0,3), ∨ OA = 1, OB = 3. ∨ CAD + ∨ Bao = 90 degree, ∨ ABO + ∨ Bao = 90 degree, ∨ CAD = ∨ AOB. ∨ AC = AB, ∨ CAD = ∨ AOB = 90 degree, ∨ ADC ≌ boa, ∨ CD = OA = 1, ad = ob = 3, ∨ od = OA + ad = 4
In the plane rectangular coordinate system, the coordinates of a, B and C are (0,1), (3,0), (2,2) respectively. Find the area of △ ABC
20
RELATED INFORMATIONS
- 1. As shown in the figure, BD is the extension of the intersection of the middle line AE ⊥ BD on the edge AC of △ ABC at the point ECF ⊥ BD. be + BF. = 2bd is proved at the point F It's sloppy
- 2. As shown in the figure, in △ ABC, BD and CD divide ∠ ABC and ∠ ACB equally, and try to explain ∠ d = 90 ° + & # 189; ∠ a
- 3. In the triangle ABC, BD.CD They are the diplomatic bisector of angle ABC and angle ACB
- 4. In triangle ABC, AF: FB = BD: DC = Ce: AE = 3:2, be, CF and ad intersect point igh on BC, AC and ab respectively, and the area of IgH is 1
- 5. As shown in the figure, in △ ABC, ∠ B = 90 °, ab = 3, AC = 5, fold △ ABC so that point C coincides with point a, and the crease is De, then the perimeter of △ Abe is______ ．
- 6. It is known that: as shown in the figure, point D is a point on the edge AC of △ ABC, passing point D makes de ⊥ AB, DF ⊥ BC, e and F are perpendicular feet, then passing point D makes DG ∥ AB, intersecting BC at point G, and de = DF. (1) prove: DG = BG; (2) prove: BD bisects EF vertically
- 7. RT, in triangle ABC, ab = AC = 10, point D on BC, de parallel AC, DF parallel AB, find the perimeter of quadrilateral AEDF
- 8. As shown in the figure, in △ ABC, de ‖ BC, ad: DB = 3:2 (1) find the value of DEBC; (2) find the value of bced & nbsp; of s △ ADEs quadrilateral
- 9. As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 4cm, BC = 3cm, now fold △ ABC to make vertex A and B coincide, then the length of crease De is () cm A. 52B. 154C. 158D. 5
- 10. Triangle ABC, AC = BC, angle c = 90 degree, point D on BC, fold triangle ACD along ad, point C just falls on point E on AB, if AB = 6cm Finding the perimeter of triangle deb The picture can't be sent out
- 11. (1 / 3) the known triangles ABC, (|) BD and CE are bisectors of angle B and angle c, respectively. Through point a, make AF vertical BD, Ag vertical CE, the perpendicular feet are f, G, connecting FG (1 / 3) the known triangles ABC, (|) BD and CE are bisectors of angle B and angle c, respectively. Through point a, make AF vertical BD, Ag vertical CE, and the perpendicular feet are f, G, connecting FG. (2) if C
- 12. It is known that in △ ABC, points D, e and F are points on BC, AB and AC respectively, AF is parallel to ED, and AF = ed, extend FD to point G, so that DG = FD, and verify that ED and Ag are equally divided,
- 13. In △ ABC, ∠ C = 90 ° and the intersection line BC of AB is at D. if ∠ bad - ∠ DAC = 22.5 °, then the degree of ∠ B is______ ．
- 14. In △ ABC, the vertical bisectors of AB and AC intersect BC at points E and f respectively. If ∠ BAC = 115 °, then ∠ EAF=______ Degree
- 15. If the inscribed circle of RT triangle is tangent to the hypotenuse AB at D, and ad = 1, BD = 2, then s ABC =? AB is the hypotenuse If the inscribed circle of RT triangle and hypotenuse AB are tangent to D, and ad = 1, BD = 2, then s ABC =? AB is hypotenuse How to use area
- 16. As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 80 °, point P is within △ ABC, ∠ PBC = 10 °, PCB = 30 °, calculate the degree of ∠ BAP What are you doing?
- 17. In △ ABC, ab = AC, ∠ BAC = 80 ° and P is a point in △ ABC, so that the angle PBC = 10 ° and ∠ PCA = 20 °. Find the degree of ∠ PAC
- 18. As shown in the figure, we know that the side length of equilateral triangle ABC is 3, M is a point on AC, passing through point m, make me parallel AB, intersect BC at point E, make MF ⊥ AB at point F, let AF = x, the median line length of trapezoidal emfb is y, find out the function analytic formula of Y and X, and write out the value range of X
- 19. AB = ad ∠ ABC = ∠ ADC, AC and BD intersect at point O. it is proved that AC is the vertical bisector of BD
- 20. Each vertex of the triangle ABC makes a parallel line AD / / be / / FC, which intersects with the extension of BC, Ca and BA at points D, e and f respectively Prove s triangle def = s double triangle ABC