Six Concyclic Points II
What Is This About?
A Mathematical Droodle

Explanation

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A solution by Achilleas Sinefacopoulos was published next year (Am Math Monthly, 107, p. 572) and, as a theorem, it appeared in [Suceava and Yiu, p. 191].

Since BC||YZ, ∠EDC = ∠EZA. In addition, ∠EDC = ∠DEC (since CE and CD are two tangents to a circle from the same point) and ∠DEC = ∠AEZ (as vertical). We may conclude that ∠AEZ = ∠EZA so that ΔAEZ is isosceles: AE = AZ. Similarly, AF = AY. But AE and AF are the tangents from A to the incircle of ΔABC. Thus all four segments are equal:

In ΔEYZ, the median from E to YZ equals half of the latter, implying that the triangle is right: ∠YEZ = 90°. Similarly, ∠YFZ = 90°. This makes YE and ZF altitudes of ΔDYZ. Their intersection H_a is the orthocenter of that triangle. It follows that DH_a is the third altitude and, since BC||YZ, DH_a ⊥ BC. But D is the point of tangency of the incircle of ΔABC with BC, implying ID ⊥ BC and subsequently the collinearity of D, I, and H_a. Further, the right angle DEH_a (with E and D on the incircle) needs to be subtended by a diameter, placing H_a on that circle and making H_a antipodal to D: DI = IH_a.

Finally, the circle through F (the foot of an altitude), E (the foot of another altitude), and A (the midpoint of the third side) is necessarily the nine-point circle in ΔDYZ. Thus the circle passes through the midpoints E' and F' of the sides DZ and DY and the Euler point I on the altitude from D.

References

1. B. Suceava, P. Yiu, The Feuerbach Point and Euler lines, Forum Geometricorum, Volume 6 (2006) 191-197.

Nine Point Circle

• Nine Point Circle: an Elementary Proof
• Feuerbach's Theorem
• Feuerbach's Theorem: a Proof
• Four 9-Point Circles in a Quadrilateral
• Four Triangles, One Circle
• Hart Circle
• Incidence in Feuerbach's Theorem
• Six Point Circle
• Nine Point Circle
• 6 to 9 Point Circle
  • Six Concyclic Points
• Six Concyclic Points II
• Bevan's Point and Theorem

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