Given a (0,3), B (- 1,0), C (3,0), the coordinates of point D are determined to make the quadrilateral ABCD an isosceles trapezoid

First of all, it is obvious that ad is parallel to the isosceles trapezoid of BC, and the ordinate of D is the same as that of a, which is 3,
It is not difficult to get the abscissa as 2, that is, the D coordinate as (2,3)
Note that if there is an isosceles trapezoid with ab parallel to CD, let's make the vertical line L of ab. if the intersection point with the horizontal axis is on the right side of C, then the isosceles trapezoid exists,
If K (AB) = (3-0) / (0 + 1) = 3, then K (L) = - 1 / 3, and if passing through (- 1 / 2,3 / 2), then l:
(Y-3 / 2) = - 1 / 3 * (x + 1 / 2), or y = - 1 / 3 * x + 4 / 3 (1)
Let y = 0, x = 4 > 3, which means that the isosceles trapezoid exists
In this case, D and C are symmetric with respect to L. it is not difficult to obtain D (16.5,3 / 5)

It is known that the quadrilateral ABCD is a right angled trapezoid, and a (0,3) B (- 1,0) C (3,0) calculates the coordinates of vertex D

Since it is a right angle trapezoid, there must be a right angle, and there must be two right angles!
According to the three points a, B and C given in the title, we can simply draw their positions in the coordinates, connect a-b-c in turn, there is no right angle, now the only right angle should appear in the angle formed by D
According to point a, the ordinate of point D is 3, and according to point C, the abscissa of point D is 3, so the coordinate of point D is (3,3)

In the quadrilateral ABCD, ∠ d = 60 °, the value of ∠ B is 20 ° larger than that of ∠ a, and the value of ∠ C is 2 times of that of ∠ a

Let a = x, then B = x + 20 ° and C = 2x. The theorem of sum of internal angles of quadrilateral obtains x + (x + 20 °) + 2x + 60 ° and 360 ° respectively, and the solution is x = 70 °. A = 70 °, B = 90 ° and C = 140 °

Given a (0,3), B (- 1,0), C (3,0), the coordinates of point D are determined to make the quadrilateral ABCD an isosceles trapezoid