It is known that in the parallelogram ABCD, if the absolute value of the vector AB is 4 and the absolute value of the vector ad is 5, what is the inner product of the vector AC and the vector BD?

It is known that in the parallelogram ABCD, if the absolute value of the vector AB is 4 and the absolute value of the vector ad is 5, what is the inner product of the vector AC and the vector BD?

（1） If vector ad = (3,5), find the coordinates of point C. (2) when the absolute value of vector AB = the absolute value of vector ad, find the help of point P.. (1) vector AC = vector AD + vector AB = (9,5) C (10,6) (2) at this time, the parallelogram is a diamond, vector BP = 1 / 3 * (vector Ba + vector BC) BC = 6, let C (√ 6cosx, √ 6

Let a triangle ABCD have a vector DC = 1 / 2, a vector AB and an absolute value of a vector ad = 1 / 2Ab, and an absolute value of a vector ad = an absolute value of a vector BC

DC = AB / 2, that is: DC ‖ AB, and: 2 | DC | = | ab|
Also: | ad | = | BC | = | ab | / 2
So: | ad | = | BC | = | DC|
The quadrilateral ABCD is an isosceles trapezoid

The edges and diagonals of space quadrilateral ABCD are a, AB, and the midpoint of CD is m, n. It is proved that Mn is perpendicular to AB and Mn is perpendicular to CD

Proof: because m is the middle point of AB, and each side is equal
It can be concluded that in triangle ABC and triangle abd, AB is perpendicular to MC and ab is perpendicular to MD
So there is: ab vertical plane MCD
Mn and CD belong to planar MCD
So: Mn vertical AB, Mn vertical CD

As shown in the figure, in the quadrilateral ABCD, ∠ C = ∠ B = 90 °, m and N are the midpoint of the diagonal AB and CD respectively, connecting Mn. What is the special position relationship between Mn and CD?

∵ Mn = 8 / 8ec = 8 / 8 (bc-ec) MC / / DC ∵ AB = de ∵ Mn = 8 / 8 (dc-ab)