## Bisector Parabola

Consider the excenter I of triangle ABC (intersection point of inner bisector at A and exterior bisectors at B, C). Take an arbitrary point F of side BC and reflect it on the (exterior) bisectors BI, CI to obtain points G, H.
[1] The sum of the signed lengths (measuring from A) of AG+AH is equal to the perimeter s of ABC.
[2] angle(GIH) = pi-A and the circumcircle of triangle AGH passes through I.
[3] Extend sides AB, AC to the isosceles AED with length of equal sides equal to the perimeter s of ABC. Then AH is equal in length with GE.
[4] The middle J of GH lies on line MN, joining the middles M, N of AE and AD.
[5] Line GH envelopes a parabola touching sides AB, AC at E and D. The parabola touches also MN at its middle K, which is its vertex. The focus of the parabola is point I and its directrix is parallel to MN at distance p (parameter of the parabola) equal to s*sin2(A/2).

[1] Obvious since BG = BF = BC+CH.
[2] angle(GIH) = angle(GIB)+angle(BIH) = angle( BIF)+angle(BIC)-angle(HIC) = angle(BIF)+angle(BIC)-angle(CIF) = angle(BIC)+(angle(BIF)-angle(CIF)) = 2*angle(BIC) = pi-A (since angle(BIC) = (pi-A)/2).
[3] By [1] GA+AH = s and the same is true for AG+GE.
[4] Follows easily from [3]. Notice that M, N are the touch-points with the sides of the excircle centered at I. It is a well known property that AM = AN = s/2 (see Bisector1.html ).
[5] Follows from the characteristic property of tangents to a parabola: The symmetric of a point I (focus) with respect to the tangents lies on a line (directrix). All the properties referred follow easily from the previous remarks.
Notice the location of the orthocenter L of triangle AGH on the directrix (see TrianglesCircumscribingParabolas.html ), due to the fact that all sides of AGH are tangents of the parabola. For the same reason the circumcircle of AGH passes through the focus I.