## Parabolas from equal segments

Given a triangle ABC select on the oriented side-line BA segment BP of given oriented length x. On the other side-line CA set points {A1,A2} at coordinates {x,-x} correspondingly starting from C. The envelopes of the two lines {PA1,PA2} are two parabolas having for axes the two bisectors of the angle A.

[1] Consider the second case in which {P,A2} are on different sides of BC, BP=CA2. Then in parallelogram APP1A2 side AA2=AC-CA2, AP=c+PB. Thus, adding we get AA2+AP=b+c and vertex P1 of the parallelogram moves on a fixed line A'A'', which is the basis of an isosceles with sides AA'=AA''=b+c.
Thus line PA2 is the diagonal of a parallelogram APP1A2 of which vertex P1 moves on the basis of the fixed isosceles triangle AA'A''. This is the typical generation method of an Artzt parabola (see Artzt_Generation.html ).
The parabola is symmetric with respect to the bisector of angle A and has its vertex at the middle M1 of the line joinining the middles of AA' and AA''.
[2] Analogous is the reasoning for the first case in which points {P,A1} lie on the same side of BC and BP=CA1. There is an analogous isosceles AB'B'' with sides AB'=AB''= c-b and the corresponding description of the envelope as an Artzt parabola, envelopped by the diagonals of parallelograms APP2A1.

Remark Both parabolas are inscribed in the triangle ABC. As noticed in [12] of ParabolaSkew.html the focus of parabolas inscribed in a triangle lies on the circumcircle of the triangle. Thus, the two foci {F1,F2} are the two intersection points of the two bisectors at A with the circumcircle of ABC.