Consider a parabola (c) and the triangle formed by two tangents AB, AC and their chord of contacts BC. Line FE joining the middles of AB, AC is also tangent to (c) at its middle K. This is one (of three) Artzt parabola for ABC. To find the intersection points of (c) with the inner Steiner ellipse (touching the triangle at the middles of the sides). The answer is: on the parallel to BC from the centroid at distance BC/(2*sqrt(3)) from it.
The proof can be reduced to a canonical parabola. Namely the analogous parabola when ABC is equilateral (shown below). In that case the Steiner ellipse is the incircle of the triangle and the intersection with the parabola is easily calculated to occure at these points. The proof is then transfered to the general case via an affinity that maps the vertices of the equilateral to corresponding vertices of the general triangle. See ParabolaTrapezium.html for another case of application of this method.
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