In the parallelogram ABCD, point m is the midpoint of AB, point n is on BD and BN = 1 / 3bd. It is proved that m, N and C are collinear 0 Use vectors if you can

It can be seen that: △ BMN ∽ CDN
Then, BMN = NCD and Mn / NC = 1 / 2
It is known that m, N and C are collinear

In the parallelogram ABCD, M is the midpoint of AB, n is the point on BD, BN = 1 / 3bd

Let AB vector be a
The Ad vector is B
So AC = AB + ad = a + B
BD=BA+AD=b-a
Vector Mn = MB + BN = AB / 2 + BD / 3 = A / 2 + (B-A) / 3 = A / 6 + B / 3
Vector MC = ma + AC = - AB / 2 + AC = - A / 2 + A + B = A / 2 + B
So MC = 3MN
So MC Mn
And because MC and Mn all pass the m-point,
So MNC three points are collinear

In the parallelogram ABCD, vector ad = vector a, vector AB = vector B, M is the midpoint n of ab. on DB, DN = 2Nb, it is proved that three points of MNC are collinear

MN=MB+BN=1/2AB+1/3BD=1/2AB+1/3(BA+AD)=1/6b+1/3a
CN=CB+BN=-AD+1/3BD=-AD+1/3(BA+AD)=-2/3a+1/3b
So Mn = - 1 / 2cn
So MNC three points are collinear

Given that P is any point in the cube abcd-a1b1c1d1 (including the surface of cube) with edge length 1, the maximum product of vector AP and vector AC is

The space rectangular coordinate system is established with a as the origin, AB as X axis, ad as y axis, Aa1 as Z axis
Then a (0,0,0), C (1,1,0)
Let P (x, y, z), then x, y, Z ∈ [0,1]
The vector AP = (x, y, z), the vector AC = (1,1,0)
The vector AP * AC = x + y
∵ x, y can take their maximum value 1 at the same time
When x = 1 and y = 1, the vector AP * AC has the maximum value x + y = 2
That is to say, the maximum product of vector AP and vector AC is 2

In the parallelogram ABCD, point m is the midpoint of AB, point n is on BD and BN = 1 / 3bd. It is proved that m, N and C are collinear 0 Use vectors if you can