Circumscribed points

To construct a triangle, given the intersection points of the circumcircle and :
- altitudes
- or angle bisectors
- or medians


The given points are the reflections of orthocenter through the sides.
The orthic triangle of ABC is therefore (line of midpoints) homothetic of the given triangle XaXbXc in ratio 1/2, through center H.
OA being perpendicular to the side of the orthic triangle is also perpendicular to XbXc, hence the construction.
The perpendicular from O to XbXc intersects the circumcircle in A and A'
The perpendicular from O to XaXc intersects the circumcircle in B and B'
and the 4 solutions ABC, AB'C', A'BC', A'B'C
For each of them, the orthocenter is the intersection of altitudes AXa and BXb, then C is the other intersection point of HXc with the circumcircle.
This orthocenter is the center of an incircle or excircle to XaXbXc.
Other method : The vertices A,B,C are the midpoints of arcs XbXc, XcXa, XaXb.

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Angle bisectors

Sorry, your browser is not Java compliant. The given points are the midpoints of the segments connecting the incenter to excenters : Xb midpoint of IJb and Xc midpoint of IJc.
XbXc is therefore parallel (line of midpoints) to JbJc. Hence perpendicular to angle bisector AXa.
Hence the construction : the perpendicular to XbXc from Xa intersects again the circle in A etc.

The intersect points of external angle bisectors with the circumcircle are the points diametrally opposite to the given points, line JbJc is the parallel to YbYc through Ya.
This gives the variant when the given points are intersections Ya,Yb,Yc of external angle bisectors with the circumcircle.


This is the tough question...

We may prove that the solutions are given from the foci of the Steiner ellipse in triangle A'B'C', tangent to sides of A'B'C' in their midpoint.
These foci G1 and G2 are the centroids in the solution triangles A1B1C1 and A2B2C2.
"Just" to construct the foci of this ellipse, with classical methods about conic sections.


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