As you see the point of concurrency in question, P, is on the Euler Line
for the triangle XYZ. There exist many interesting ratios on this line.
This point of concurrency, P, is the midpoint of Euler's line in triangle XYZ. This means that the center of the nine point circle which circumscribes triangle XYZ is the point P.
The point labeled NP is the center of the nine point circle. The other
P is the label (for the same point) which is the point of concurrency in
When we construct a line through one of the vertices, say A, and find
the locus of point P, we see that it seems to be an ellipse but it never
actually closes. If you could finish drawing the ellipse, it would pass
through two vertices (in this case, A and B) of the original triangle and
the centroid (in this case, C') of the equilateral triangle made of the
side containing those two vertices. Sometimes the "ellipse" will
go through the third vertex of the original triangle also (C), that is,
when angle ACB is big enough.
At a certain point where an angle is around 120 degrees, the ellipse
changes to a hyperbola, which seems to always pass through two vertices.
The interesting ratios are still there just by nature of the Euler line
Looking at a trace of P in this situation, we see what appears to be
another open ellipse, going through at least two vertices. (Triangle A'B'C'
is still green, the squares on each side are hidden in the interest of space.)
Repeat the investigation but let A', B', and C' be the external vertices
of the externally constructed equilateral triangles.
If you construct a square on each side of ABC, then the angle BAC' will
be 45 degrees. If C' is the external vertex of an equilateral triangle,
angle BAC' will be 60 degrees.
That center point (not P but the A', B', or C') is just located at different points along the perpendicular bisector. Rotate a side from 0 degrees to 90 degrees around a fixed vertex and the intersection of that side and the perpendicular bisector will be A', B', or C'.
The angles L, M, and N are the ones formed when the sides are rotated.
The pink dotted lines are the perpendicular bisectors, so the point of intersection
shown here is the circumcenter.
Click here to explore the construction.
Click here to see the locus traces.
Drag any vertex of the original triangle.
Isn't that cool?