All triangles are made of three points (called vertices), and the lines connecting them. Now image three random non-collinear points (non-collinear means that all three points are not in a straight line). Call these three points \(P_1\), \(P_2\) and \(P_3\):
Let the line \({\color{blue} L_1}\) be a line going from \(P_1\) to \(P_2\). This is one side of the triangle:
Let the line \({\color{green}B_1}\) be the locus of points equidistant from \(P_1\) and \(P_2\):
Line \({\color{green}B_1}\) is the perpendicular bisector of \({\color{blue} L_1}\), and all points in \({\color{green}B_1}\) are equidistant from \(P_1\) and \(P_2\). Now let the line \({\color{blue} L_2}\) be a line going from \(P_2\) to \(P_3\). This is another side of the triangle:
Let the line \({\color{green}B_2}\) be the locus of points equidistant from \(P_2\) and \(P_3\):
Line \({\color{green}B_2}\) is the perpendicular bisector of \({\color{blue} L_2}\), and all points in \({\color{green}B_1}\) are equidistant from \(P_2\) and \(P_3\). By the definition of a triangle, the angle between \({\color{blue} L_1}\) and \({\color{blue} L_2}\) cannot be 180°. That means the bisectors \({\color{green}B_1}\) and \({\color{green}B_2}\) are definitely not parallel, which also means they must intersect at one point. Let's call this intersection point \(C\):
All points in \({\color{green}B_1}\) are equidistant from \(P_1\) and \(P_2\), and all points in \({\color{green}B_2}\) are equidistant from \(P_2\) and \(P_3\). This means the intersection \(C\) is equidistant from \(P_1\), \(P_2\) and \(P_3\):
If the distance between \(P_1\) and \(C\) is \(r\), then the distance between \(P_2\) and \(C\), and the distance between \(P_3\) and \(C\) would also be \(r\). This means if you make a circle around \(C\) with radius \(r\), it would pass through the points \(P_1\), \(P_2\) and \(P_3\):
Since you can do this with any three points, this proves that every triangle can be circumscribed by a circle.