It is known that in the quadrilateral ABCD, ab = ad = 4, BC = 6, CD = 2,3 vector AB * vector AD + 4 vector CB * vector CD = 0, we can find the circumcircle radius r of triangle ABC

It is known that in the quadrilateral ABCD, ab = ad = 4, BC = 6, CD = 2,3 vector AB * vector AD + 4 vector CB * vector CD = 0, we can find the circumcircle radius r of triangle ABC

AB = ad = 4, BC = 6, CD = 2,3 vector AB * vector AD + 4 vector CB * vector CD = 0,
That is, 3 * 4 * 4cosa + 4 * 6 * 2cosc = 0, that is, cosa + COSC = 0, so cosa = - COSC,
Because 0

In ABCD, G is the center of gravity of △ ABC, be = 2ed, with {AB, AC, ad} as the base, then Ge=

The ratio of the distance from the center of gravity to the vertex and the distance from the center of gravity to the midpoint of the opposite side is 2:1
The relationship between BG and AB, BC can be obtained
ge=gb+be
be=(1/2)(ba+ad)

Proposition: a quadrilateral ABCD is a parallelogram if and only if the vector AB = the vector DC. That's right. Why?

The vector AB is equal to the vector DC, which indicates that they are equal in length and direction. The same direction further indicates that they are parallel, and the quadrilateral with the same size and parallel opposite side is a parallelogram, so the proposition is correct

The following propositions: 1. If and only if two vectors are equal, they have the same starting point and the same ending point; 2. If AB = DC, then ABCD is a parallelogram
3. If the quadrilateral ABCD is a parallelogram, then AB = DC; 4. A = B, B = C, then a = C, where the correct sequence number is

It's three and four
1. The vectors are still equal after translation
2. ABCD can be in the same line