Given three points X, Y, Z, show how to construct a triangle ABC which has circumcenter X, Y the midpoint of BC and BZ an altitude.
XY is the perpendicular bisector of BY, so BC must be the line L through Y perpendicular to XY. Now ∠BZC = 90o, so Z lies on the circle diameter BC. Hence YZ = YB = YC. So the circle center Y radius YZ meets L at B and C. Now take the circle center X radius XB. The line CZ must intersect it again at A.