In trapezoidal ABCD, ad is parallel to BC, angle ABC + angle c = 90 degrees, ab = 6, CD = 8, MNP is the midpoint of AD, BC and BD respectively, what is the length of Mn

In trapezoidal ABCD, ad is parallel to BC, angle ABC + angle c = 90 degrees, ab = 6, CD = 8, MNP is the midpoint of AD, BC and BD respectively, what is the length of Mn

Make AE ∥ DC through point a and hand BC to point E
Connect MP, PN, Mn
∵∠ ABC + ∠ C = 90 degrees
The angle between ABC and AEB is 90 degrees
The angle of Bae is 90 degrees
MNP is the midpoint of AD, BC and BD
The MP and PN are the median lines of abd and BDC, respectively
∴AB‖MP PN‖DC PM=1/2AB=3 PN=1/2DC=4
Also known as ∵ AE ∥ DC
∴PN‖AE
The angle is equal to or complementary to each other if the two sides of one angle are parallel to each other
The triangle PMN is a right triangle
According to Pythagorean theorem, Mn = 5

As shown in the figure, in ladder ABCD, ad ‖ BC, ∠ B + ∠ C = 90 °, ab = 6, CD = 8, m and N are the midpoint of AD and BC respectively, then Mn is equal to ()
A. 4B. 5C. 6D. 7

As shown in the figure: through the point m, make me ‖ AB, MF ‖ CD, ∠ men = ∠ B, ∠ NFM = ∠ C, ∫ B + ∠ C = 90 degree, ∫ MEF + ∠ MFE = 90 degree, ∫ EMF = 90 degree. ∫ ad ‖ BC, ∫ me = AB = 6, MF = CD = 8, am = DM, BN = CN. ∫ EF = 10, en = FN. ∫ Mn = 12ef = 5

If E1 and E2 are two unit vectors perpendicular to each other in the plane, and a = E1 + E2, the vector parallel to a can be expressed as
A a/2
B ±a/2
C A / radical 2
D ± A / root 2

It is concluded that the unit vector parallel to a vector is ± A / | a |
Let's prove that:
[certification]
Let the unit vector which is collinear with vector a (x, y) be A0 = (x0, Y0),
Unit vector is the vector with module 1, and module 1 is the root sign (x0 ^ 2 + Y0 ^ 2) = 1
That is, x0 ^ 2 + Y0 ^ 2 = 1 ①
Vector A0 and vector a are collinear, so Y0 / y = x0 / X,
That is, Y0 = x0, Y / X ②
Substituting ② into ①, we can get: x0 ^ 2 + x0 ^ 2 y ^ 2 / x ^ 2 = 1,
x0^2*(( x^2 +y^2) /x^2)=1,
x0^2* =x^2/( x^2 +y^2),
The solution is x0 = ± X / √ (x ^ 2 + y ^ 2),
By substituting x0 into ②, Y0 = ± Y / √ (x ^ 2 + y ^ 2) can be obtained
Where √ (x ^ 2 + y ^ 2) is the module of vector a (x, y)
The unit vector collinear with a is ± A / | a|
Given the vector a = (1,1), | a | = √ 2,
Then the unit vector collinear with vector a is (√ 2 / 2, √ 2 / 2) or (- √ 2 / 2, √ 2 / 2)
If E1 and E2 are two unit vectors perpendicular to each other in the plane and a = E1 + E2, the unit vectors parallel to a can be expressed as
A a/2
B ±a/2
C A / radical 2
D ± A / root 2
[interpretation]
Because a = E1 + E2,
So | a | ^ 2 = (E1 + E2) ^ 2
=e1^2+e^2+2e1•e2
=1+1+0
=2,
So | a | = √ 2,
The unit vector parallel to a can be expressed as ± A / √ 2
Choose D
[example] what are the coordinates of the unit vector parallel to the vector a = (- 12,5)?
The unit vector parallel to a vector is ± A / | a|
Because | a | = 13
So the answer is ± (- 12,5) / 13
That is (- 12 / 13,5 / 13) or (12 / 13, - 5 / 13)

In △ ABC, D is the point on BC, and BD = 1 / 2dc, e is the point on ad, and AE = 2ed. If vector AB = vector E1, vector AC = vector E2, E1 and E2 are used to represent vector CE

Because vector AE = 2, vector ed, vector BD = 1 / 2, vector DC
Vector ed = 1 / 3 vector ad, vector CD = 2 / 3 vector CB
Vector CB = vector ab - vector AC
Vector ad = vector AC + vector CD = vector AC + 2 / 3 vector CB = vector AC + 2 / 3 (vector AB vector AC) = 2 / 3 vector AB + 1 / 3 vector AC
Vector CE = vector CD vector ed = 2 / 3 (vector AB vector AC) - 1 / 3 (2 / 3 vector AB + 1 / 3 vector AC) = 2 / 9 vector Ab-5 / 9 vector AC = 2 / 9 vector e1-5 / 9 vector E2