Any line passing through the center of gravity of △ ABC intersects AB and AC at points D and e respectively AD＝x AB， AE＝y AC, XY ≠ 0, then 1 X+1 The value of Y is______ .

0

For the convenience of narration, the following words are omitted for the convenience of narration. The word "vector" is omitted below for convenience of narration. Point D is used to do DH ∥ AC, and DH is to be in H. then △ BDH ≔ BAC, the similarity ratio is | BD | / | BA | = 1 -

It is known that ad is a midline of the triangle ABC, point E is on edge AC, and vector AE = 1 / 4, vector AC, ad and be intersect at point O, if vector AB and BC are the basis It is known that ad is a midline of triangle ABC, point E is on edge AC, and vector AE = 1 / 4, vector AC, ad and be intersect with point O. if vector AB and BC are the basis, vector Ao can be expressed as X vector AB + y vector BC (x, y belongs to R), then the values of real numbers x and y are (?) A.1/4,1/2 B.1/2,1/4 C.1/5,2/5 D.2/5,1/5

Vector Ao = AB + Bo = AB + MBE (because the vector Bo and be are collinear, so Bo = MBE)
= AB+m(AE-AB)
= AB+m(1/4AC -AB)
= AB+m[1/4(AB+BC) –AB]
=(1-3m/4) AB+m /4 BC.
Because the vector Ao is collinear with AD,
So the vector Ao = NAD
=n(AB+BD)
=n(AB+1/2BC)
= n AB+ n /2BC
To sum up, the vector Ao = (1-3m / 4) AB + m / 4 BC = n AB + n / 2BC
∴1-3m/4= n,m /4 =n /2,
M = 4 / 5, n = 2 / 5
The vector Ao = N2 / 5ab + 1 / 2 / 5BC
Choose D

If the vector ABC / 5 is the area of the triangle in the plane of ABC + 5, let ABP be the area of the triangle

Let the angle a * a * sin B = 0 * a * sin B = 0 * a * 2, then let a * a * a = a * a * a + A * b = 0 * a * a + b * a = 0 * a * a + b * a = 0 * a * a + b * a + A * a + b * a = 0 * a * a + b * a + A * a + b * a = 0 * a * a + b * a + A * b = 0 * a, b * a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, a, B, B, a, B, B, a, B, a, B, a, B, a, B, B, a, B, B, a, B

Let p be a point in the triangle ABC, and the vector AP = 2.5 vector AB + 1.5 vector AC. the area ratio of triangle PBC to triangle ABC is

Let ad = tap = 2T / 5ab + T / 5AC, so 2T / 5 + T / 5 = 1, t = 5 / 3, the area ratio of triangle PBC to triangle ABC = the height ratio of triangle PBC to triangle ABC = Pd / ad = 2 / 5

In triangle ABC, vector am = 1 / 3 vector AB, vector an = 1 / 4 vector AC. try to use vector a and vector B to represent vector AP

There are two methods
Method 1
Extend BF to CD extension line at P
∵ ab ∫ CD, ᙽ AB / DP = AF / DF = 1 / 3, then AB / CP = 1 / 4
∵BE=2AB/3,∴BE/CP=2/3×1/4=1/6
∴EG/CG=BE/CP=1/6,EG/EC=1/7
Vector eg = (1 / 7) vector EC = (1 / 7) (vector EB + vector BC) = (1 / 7) (2A / 3 + b)
ν vector Ag = vector AE + vector eg = A / 3 + (1 / 7) (2A / 3 + b) = 3A / 7 + B / 7
Method 2

In the triangle ABC, D and E are the midpoint of BC and AC respectively, f is the point above AB, and vector AB = 4 vector AF. if vector ad = x vector AF + y vector AE, then x =?, y =?

The following is the full index vector
AD=1/2(AB+AC)=1/2(4AF+2AE)=2AF+AE
So x = 2, y = 1

In △ ABC, am: ab = 1:3, an: AC = 1:4, BN and cm intersect at point E, vector AB = a, vector AC = B, and vector AE is represented by a, B

In △ ABC, it is known that D is a point on the edge of ab. if vector ad = 2 vector dB, vector CD = 1 / 3 vector Ca + λ vector CB, then λ = () A：2/3 B：1/3 C：-1/3 D：-2/3

Change the following λ into the one in your topic. From vector CD = 1 / 3 vector Ca + λ vector CB, we can get (λ - 2 / 3) vector CA = λ vector ab - vector ad because vector ad = 2 vector dB, we get (λ - 2 / 3)

In the triangle ABC, let d be a point on the edge of ab. if vector ad = 3 * vector dB, vector CD = 1 / 4 * vector Ca + X * vector CB, find the value of X The answer is 3 / 4. Why?

prove:
In fact, this is the formula of fixed score point, which can be proved again,
Vector CD
=Vector Ca + vector ad
=Vector Ca + (3 / 4) vector ab
=Vector Ca + (3 / 4) (vector CB vector CA)
=(1 / 4) vector Ca + (3 / 4) vector CB

Any line passing through the center of gravity of △ ABC intersects AB and AC at points D and e respectively AD＝x AB， AE＝y AC, XY ≠ 0, then 1 X+1 The value of Y is______ .