We will not consider edge cases in this problem for brevity since they don't impact the solution that much. Call the points $A,B,C$. The idea is to show that there exists a line that fixes two of the points and moves the third after folding. The following lemma is the heart of the problem. Let $X,Y,P,Q$ be four points not all collinear. Then the perpendicular bisector of $PQ$ intersects segment $XY$ iff $XP-XQ$ and $YP-YQ$ have different signs. Proof: Define $f(\lambda)=MP-MQ$, where $M=\lambda X+(1-\lambda)Y$. Clearly $f$ is continuous, so by IVT there's a point that's equidistant from $P$ and $Q$, an intersection point. The converse is also easy. We first deal with the case where $A,B,C$ are collinear. Assume that $B$ lies between $A$ and $C$ with $BC>BA$. Construct equilateral triangle $\triangle BCD$. Then $BA-BD$ and $CA-CD$ are both positive. Folding over the perpendicular bisector of $AD$ fixes $B$ and $C$ and $A\mapsto D$. Now we consider a triangle. Construct equilateral triangles $\triangle BCD,\triangle CAE,\triangle ABF$ on the interior of $\triangle ABC$. Let the perpendicular bisectors of $AD,BE,CF$ be $m_a,m_b,m_c$. The claim is that at least two of $m_a,m_b,m_c$, do not intersect their respective sides. Assume that $m_a$ intersects segment $BC$. Then by the lemma, $BA-BD=c-a$ and $CA-CD=b-a$ have different signs. WLOG let $b>a>c$. Since $b-c$ and $a-c$ have the same sign as well as the pair $c-b,a-b$, $m_b$ and $m_c$ do not intersect segments $CA$ and $AB$. However, this isn't enough to solve yet. We still need to show at least one of $m_b$ and $m_c$ separates its respective vertex from the other two. Fortunately, this is easy. Since $AF<AC$ and $BF<BC$ from $b>a>c$, $A,B,F$ are on the same side of $m_c$, while $C$ is on the other. Thus, folding the paper over $m_c$ will do the trick.