$\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{Generalization}$ $$\definecolor{A}{RGB}{80,0,200}\color{A} \text{If } \triangle ACE\equiv \triangle DFB\text{ in the same orientation and they aren't translation of one another,}$$$$\definecolor{A}{RGB}{80,0,200}\color{A}\text{then lines connecting the midpoints of segments in pairs }(AB, DE),\ (BC,EF),\ (CD,FA)$$$$\definecolor{A}{RGB}{80,0,200}\color{A} \text{ intersect in the midpoint of segment joining the circumcenters}.$$$\definecolor{A}{RGB}{190,0,60}\color{A}\fbox{Proof}$ Set vertices of $\triangle ACE$ $$a,c,e\in\mathbb{C}\wedge |a|=|c|=|e|=1.$$There exist $$k,l\in\mathbb{C}\wedge |k|=1\wedge k\neq 1$$such that the vertices of $\triangle DFB$ are $$d=ka+l,\ f=kc+l,\ b=ke+l.$$Because of symmetry it's sufficient to prove that points$$\frac{a+b}{2},\ \frac{l}{2},\ \frac{d+e}{2}$$are collinear. $$\frac{a+b-l}{a+b-d-e}=\frac{a+ke}{a+ke-ka-e}=\frac{a+ke}{(k-1)(e-a)}\in\mathbb{R}.\blacksquare$$#1749