Prove that three points are collinear In trapezoidal ABCD, ad parallel BC angle B = 40 degree, C = 50 degree M. N is the midpoint of BC ad Extending Ba CD intersection point P Prove that p m n three points are collinear

Prove that three points are collinear In trapezoidal ABCD, ad parallel BC angle B = 40 degree, C = 50 degree M. N is the midpoint of BC ad Extending Ba CD intersection point P Prove that p m n three points are collinear

Extending PN to BC in e
Because AD / / BC
Pan PBE is similar to PDN PCE
So an; be = PN; PE DN; CE = PN; PE
Because an = nd
So be = CE, so e is BC, and the midpoint coincides with M
So p m n three points are collinear

Given the vector M = (1,1), the angle between the vector N and the vector m is 3 π / 4, and the vector M. the vector n = - 1
Let vector a (1,0), vector b = (cosx, 2cos2 (π / 3 - X / 2)), where 0 < x < 2 π / 3, if vector n multiplies vector a = 0, try to find the value range of | vector n + vector B |

Let n (x, y) be a vector
Mn = - 1, so x + y = - 1
mn=｜m｜｜n｜cosa=√2*√（x^2+y^2)*cos3/4π=-1
That is, x ^ 2 + y ^ 2 = 1... (2)
(1) If x = 0 or x = - 1, then y = - 1 or 0
So the vector n = (0, - 1) or (- 1,0)
The angle between vector N and vector q = (1,0) is π / 2, so the direction of n is y-axis, and the modulus value is 1, so n = (0, - 1);
From 2B = a + C, we know B = π / 3, a + C = 2 π / 3.0

It is known that plane α passes through a (0,0,1), B (3,0,0), and the dihedral angle with plane xoy is 60 ° to find a normal vector of plane α
Please give a detailed process, thank you (if there is, please give an example document, thank you again)

Solution: plane equation, l for lamda
(0x+0y+1z)+L(3x+0y+0z)=0
That is Z + 3lx = 0. Its normal vector = (3L, 0,1)
The normal vector of the plane of xoy is (0,0,1)
cos(60)=1/2
So 0.5 ^ 2 = 1 / 4=
(a1a2+b1b2+c1c2)^2/(a1^2+a2^2+a3^2)(b1^2+b2^2+b3^2)
therefore
1/4=1/(9L^2+1)*1
So 9L ^ 2 = 3
L^2=1/3
So l = 1 / 3 of the positive and negative root sign
So the two normal vectors of the plane
(radical 3,0,1), (negative radical 3,0,1)