Dear Alex,

I would like to contribute here another proof I think interesting.
We use here one simple lemma:
Given two circles (O1), (O2) intersect each other at two points T, U. Any line L passing one intersection point, for example T and intersects (O1), (O2) again at V, W. Suppose Z is midpoint of VW. The result: Z is on circle (O) centered at midpoint O of O1O2 passing T, U.
It is can be proved by draw three perpendiculars from O, O1, O2 to L.
We can apply this lemma to "two equilateral triangles" by drawing two circles (O1), (O2) as circumcircles of two equilateral triangles. After that, all our seven points are midpoints of one line passing one of two intesections of (O1), (O2). Midpoints of XB, YB are two special cases when one end of segment is also intersection point.

Best regards,
Bui Quang Tuan