Construct two orthogonal circles. Begin with circle C1 centered at A with radius AB. Now, construct a tangent line, t, to C1 at the point B. Choose any point Q along t and construct circle C2 with radius QB. Now, we know that QB is tangent to circle C1 and AB is tangent to C2. That is, AB is perpendicular to QB. The circles also intersect at another point, R. Notice, the triangles QBA and QRA are congruent by the SSS triangle congruence criterion. Thus, angle R must also be a right angle. That is, QR is tangent to C1 and AR is tangent to C2. Now, consider the quadrilateral QBAR. Notice, that angles B and R are non-adjacent and supplementary ([math]90^\circ+90^\circ=180^\circ[/math]). Because the sum of interior angles of a quadrilateral is [math]360^\circ[/math], this means the other two non-adjacent angles are supplementary as well. By Theorem 4.3, this means that the quadrilateral QBAR is cyclic.