## Cevians bisectors

Here we study an interesting configuration resulting from the cevians of a point P with respect to a triangle ABC. Assume that the cevians intersect again the circumcircle (c) of the triangle at points A*, B*, C* (triangle A*B*C* is called circumcevian).
[1] Draw the bisectors of the angles of the cevians at P. For points P inside the circumcircle these lines intersect the circumcircle at six points: B'B'', B' between A and C*, C'C'' with C' between B and A* etc..
[2] From A', A'' draw orthogonals to A'A''. From B', B'' draw orthogonals to B'B'' etc.. The six lines thus defined build a symmetric, with respect to the circumcenter O, hexagon DEFGHI.
[3] The tangent conic (d) to the hexagon has one focus on P, the other being symmetric with respect to O.
[4] The tangent conic (d) is also tangent to the circumcircle (c), the touch points being on line PO.

To prove [2] notice that opposite sides of this hexagon, as for example DI and GF are orthogonal to the same line C'C'', thus parallel. Then notice that two adjacent sides and their opposites, as for example {DI, DE} and {FG, GH} are parts of two cyclic quadrilaterals C''DB'P and B''GC'P. Obviously triangles B'C''P and C'B''P are similar and from this follows easily that the quadrilaterals themselves are similar. Extend then diameter PD to intersect C'B''. These two lines are orthogonally intersected. In fact, angle(C'B''G) = angle(C'PG) = angle(B'PD). In the first and third equal angles side B'B'' is orthogonal to B''G, hence the other side PD must be orthogonal to B''C'.
An immediate implication of these facts is that the quadrangle formed by the centers of circumcircles of the two quadrangles and points P and O is a parallelogram. This implies in turn that DG contains O and is bisected by it, thus, proving the symmetry of the hexagon.
To show [3, 4] notice first that the tangent conic will be symmetric with respect to O, since the hexagon itself is so. Take then the symmetric P' of P w.r. to O and reflect PP' on one side of the tangent hexagon, for example on GH, obtaining the isosceles trapezium PP'P''Q. By the symmetry of the figure the middle O' of P'P'' will coincide with the intersection of circle (c) with the side GH of the hexagon. Besides the intersection point Q0 of PP'' with GH will coincide with the touch point of GH with the circle. Since OO' is twice the distance PP'' it follows that PQ0+Q0P' = 2*OO' = 2*r, r being the radius of (c). Thus all touch points of the hexagon are on the ellipse with foci at P, P' and major axis 2*r. This identifies this ellipse with the inscribed one of the hexagon.

### See Also

Circumcevian.html
Ellipse.html
Symmetric_hexagons.html

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