In the triangle ABC, D is the midpoint of BC, AE bisector angle BAC, make de perpendicular to AE at e, intersect AB at g, intersect AC extension line at h. proof: BG = ch = 1 / 2 (AB-AC)

In the triangle ABC, D is the midpoint of BC, AE bisector angle BAC, make de perpendicular to AE at e, intersect AB at g, intersect AC extension line at h. proof: BG = ch = 1 / 2 (AB-AC)

Cm ‖ AB intersects GH with m through C
∴∠B=∠DCM
∵ D is the midpoint of BC
∴BD=CD
And ∵ ∠ BDG = ∠ MDC (opposite vertex angle)
≌ bgd ≌ CMD (corner side)
∴BG=CM
∵CM‖AB
∴∠CMH=∠AGH
AG = ah
∴∠AGH=∠H
∴∠CMH=∠H
∴CM=CH
∴BG=CH
AB-AC = Ag + BG - (ah-ch)
=AG+BG-AH+CH
=AG+BG-AG+BG
=2BG
∴BG=CH=1/2（AB-AC)
Who knows the definition of function can go to discontinuous point
There are three kinds of discontinuities: ① removable discontinuities = the first kind of discontinuities, left limit = limit ≠ function value (or undefined)
② Jump discontinuity = left limit of the second kind of discontinuity ≠ right limit
③ Infinite discontinuity = the third kind of discontinuity limit does not exist (infinite or uncertain), such as: y = sin (1 / x) x = 0
It is known that the sum of the first n terms of the sequence {an} is Sn, satisfying an + Sn = 2n. (I) prove that the sequence {An-2} is an equal ratio sequence, and find out an; (II) let BN = (2-N) (An-2), find out the maximum term of {BN}
(I) prove: A1 = 1 is obtained from a1 + S1 = 2A1 = 2; an + 1 + Sn + 1 = 2 (n + 1) is obtained from an + Sn = 2n; 2An + 1-an = 2 is obtained by subtracting the two formulas, that is, 2An + 1-4 = An-2, that is, an + 1-2 = 12 (An-2) is an equal ratio sequence with the first term of A1-2 = - 1 and the common ratio of 12
If the vector groups A1, A2, --, as are linearly independent, then the necessary and sufficient condition for the linear independence of the n-dimensional vector groups B1, B2, BS is
The necessary and sufficient conditions for vector groups A1, A2, --, as to be linearly independent and vector groups B1, B2, BS to be linearly independent are
A vector group A1, A2, --, as can be expressed linearly by vector groups B1, B2, BS
B vector groups B1, B2, BS can be expressed linearly by vector groups A1, A2, --, as
C vector group A1, A2, --, as and vector group B1, B2, BS are equivalent
D vector group A1, A2, --, as and vector group B1, B2, BS have the same rank
Ask for detailed explanation
Choose D
If B1, B2... Is linearly independent, then its rank is equal to the number of vectors, that is s, then R (A1, A2...) = R (B1, B2...). Therefore, it is equivalent
If G is the center of gravity of triangle ABC and the midpoint of each side is D, e and F, then Gd (vector) + Ge (vector) + GF (vector) =?
The result is a zero vector
Let's leave out the vector and just use the letter
GA+GC=2GF
GA+GB=2GD
GB+GC=2GE
So GD + Ge + GF = GA + GB + GC
GA + HB = - 2gc
The result is a zero vector
For the function f (x) defined on R, we can prove that point a (m, n) is a symmetric point of F (x) image if and only if f (M-X) + F (M + x) = 2n
The function f (x) = ax ^ 3 + (b-2) x ^ 2 is an odd function on R, and the conditions for finding a and B are discussed. Whether there is a constant a in the interval [- 1,1] such that f (x) is greater than or equal to - x ^ 2 + 4x-2 is constant is also discussed
The necessary and sufficient condition of two-point symmetry is: let the coordinates of the symmetry point be: (x, y), then there are always two-point abscissa of symmetry: x-a, x + a ordinate: F (x-a) = f (x) + m, f (x + a) = f (x) - M. if we know these, we can prove it. Sufficient: a (m, n) is a symmetry point of F (x) image, then there are two expressions: F (M-X) = n + A, f (M + x) = n-a
Given that the sum of the first n terms of sequence an is Sn, satisfying an + Sn = 2n, note BN = 2-An, prove that BN is an equal ratio sequence, and find the sum of BN
The first n terms and BN
an+sn=2n;
a(n-1)+s(n-1)=2(n-1);
The above two formulas are subtracted;
The results show that 2an-a (n-1) = 2;
Then 2 * (2-An) = (2-A (n-1));
That is, 2 * BN = B (n-1);
Is an equal ratio sequence;
S1 = A1, A1 = 1;
Then B1 = 1;
bn=0.5^(n-1);
It is known that the three solution vectors of AX = B are A1, A2, A3. If (a1 + A2) - KA3 is the solution vector of AX = 0, what is k
Because a (a1 + a2-ka3) = 0,
So Aa1 + aa2-kaa3 = 0,
That is, B + b-kb = 0,
So (2-k) B = 0,
Then k = 2
In triangle ABC, ad intersects BC with D, be intersects AC with E, ad, be intersects g, BD: DC = 3:1, Ag = Gd, find BG: Ge
Through D as DF ‖ be, AC to F,
∵AG=DG,∴AE=EF,
∴2EG=DF,
And DF / be = CD / BC = 1 / 4,
∴BE=4DF=8EG,
∴BG=7EG,
BG：EG=7：1.
It is proved that the necessary and sufficient condition for the image of function y = f (x) defined on R to be symmetric with respect to x = a is f (x) = f (2a-x) (a belongs to R)
If the image of y = f (x) is symmetric with respect to x = a, then f (A-X) = f (a + x)
Let A-X = t, then x = A-T, a + x = 2a-t
F (T) = f (2a-t), that is, f (x) = f (2a-x)
If f (x) = f (2a-x) and x = A-T, then 2a-x = a + t
That is, f (A-T) = f (a + T)
That is, f (A-X) = f (a + x), so the image of y = f (x) is symmetric about x = a
It's over