The points $A(-6,-1), B(2,3), C(-1,4)$ are given in a Cartesian coordinate system. Determine the point $D$ such that the quadrilateral $A B C D$ is an isosceles trapezoid $(A B \| C D)$.
I. Solution: By construction, the point $D$ is to be established as the intersection of the line parallel to $AB$ (briefly $a$) passing through $C$ and the circle centered at $A$ with radius $BC$; let us follow this line of thought using the methods of coordinate geometry. The slope of $a$ from the given data is $\frac{3+1}{2+6}=\frac{1}{2}$, so the equation of $c$ is:

$$
y-4=\frac{1}{2}(x+1)
$$

On the other hand, the square of the length of the side $BC$ is $(4-3)^{2}=(-1-2)^{2}=10$, so the equation of the circle mentioned is:

$$
(x+6)^{2}+(y+1)^{2}=10
$$

Now, the coordinates of $D$ can be found from the system of equations formed by (1) and (2): $D_{1}(-5,2)$ and $D_{2}(-9,0)$. Of these, $D_{1}$ forms a "conventional" trapezoid, while $D_{2}$ completes $ABC$ into a parallelogram.

László Gyuris (Gyöngyös, Vak Bottyán g. III. o. t.)

II. Solution: Utilize the axial symmetry of the isosceles trapezoid, and construct $D$ as the reflection of $C$ over the axis, which is the perpendicular bisector of the parallel sides. The points on the axis $t$, being equidistant from $A$ and $B$, satisfy the equation

$$
(x+6)^{2}+(y+1)^{2}=(x-2)^{2}+(y-3)^{2}
$$

which, when rearranged in the usual form, is

$$
2 x+y+3=0
$$

The line through $C$ and perpendicular to $t$ is identical to $c$, so the midpoint $F$ of side $CD$, which we wish to reflect $C$ over, is given by the solution of the system of equations formed by (1) and (3): $F(-3,3)$. Now, denoting the coordinates of $D$ by $(u, v)$, from the fact that $F$ bisects $CD$:

$$
\frac{-1+u}{2}=-3 \quad \text { and } \quad \frac{4+v}{2}=3
$$

and from this, $u=-5, v=2$.