The given points are finitely many in number so therefore there exists a disk which contains them in its interior.
Namely, we are searching for the point which is at a maximal distance from the origin of the coordinate system. If we denote that distance by $R$, then the disk centered at the origin with radius $2R$ contains all the given points.
We draw all the possible lines between the $n$ points in the plane. They are also finitely many in number ($\binom{n}{2}$), so we can choose a point $A$ which doesn't lie on any of those lines and is outside the disk containing the points and the biggest angle under which every segment in the disk is seen is acute. We draw an arbitrary line through $A$ which doesn't intersect the disk. We rotate that line until it passes through a point from the given $n$ points and we denote it by $A_1$. On that line lies one of the sides of the triangle. We continue rotating the line until we get a line on which one of the given points lies and we denote it by $A_n$. On that line lies another side of the triangle. Each of the two lines pass only through $A_1$ and $A_n$ respectively since the point $A$ does not lie on any of the $\binom{n}{2}$ lines.

Let $A_i$ denote the point from the given $n$ points which is at the greatest distance from $A$ and let $d$ denote that distance. It may be that there are several points at a greatest distance from $A$, but then we pick an arbitrary one. We draw a tangent of the circle centered at $A$ and radius $d$. On that line lies the third side of the triangle.