## Generation of the Artzt parabola II

Given a triangle ABC, the Artzt parabola relative to side BC (see Artzt.html ) is the parabola tangent to sides AB, AC at points B and C respectively. Here it is shown that this parabola is the envelope of the diagonals GF of paralellograms AGEF for a point E gliding on BC.

The clue idea is to use the many similar triangles created by the figure. The figure which is complemented with the symmedian line AI and the median AD. The following facts result easily:
[1] The circumcircles of triangles AGF pass through a fixed point M on the symmedian AI.
In fact, let G be the intersection point of this circumcircle with side AB. Then triangles MGB and MFA are similar. They have angle(BGM)=angle(AFM). They have MG/GF=c/b and finally also BG/AF=c/b. First equality is due to the characteristic property of the points on the symmedian (see [3] in Symmedian_0.html ). Second because AF is parallel and equal to GE. Thus angle(GBM)=angle(MAF) is constant and M remains fixed.
[2] angle(MGF) = angle(MAF) is also fixed. Hence projecting M on GF creates a right-angled triangle MGQ, which by varying E varies remaining always similar to itself, has fixed vertex M and G gliding on line AB. Hence (see Similarly_Rotating.html ) the other vertex Q glides on a line L. This line is the tangent of the Artzt parabola at its vertex. The following remarks confirm this claim.
[3] Line L is orthogonal to the median AD.
This is seen by considering the special position M'M''M of triangle MGQ (ibid), in which M' is the intersection point of AB with L and M'' the projection of M on L. angle(AM'M'')=angle(M'MM'')=pi/2-angle(GAK) and this proves the claim.
[4] Triangles AGM and and AKF are similar.
This because angle(AMG)=angle(AFG) and angle(GAM)=angle(KAF). Later because of the symmetry on the bisector AH.
[5] Let O be the intersection of the parallel to the median from E with FG. Triangles AGK, EFO, GMO are similar to AMF.
In fact, by [3] AG/AK = AM/AF and angle(GAK)=angle(MAF). This proves the similarity of triangles GAK and MAF. By the parallelogram, triangle EFO equals triangle AGK. Also triangle GOE and AKF are equal and we have angle(MGO)=angle(GAK) and GM/AG = AK/KF = AK/GO. This proves the similarity of triangles GAK and MGO.
[6] Take the symmetric P of M with respect to Q. P moves on a line L', parallel to L at double distance from M than the distance of L from M (obvious).
[7] OP is parallel to the median and equal to OM.
This is also obvious, from the similarity of triangles GOM and EOF, by which O, P and E are on the same line.
[8] O is on the parabola with focus M and directrix L'. GF is the tangent to that parabola at O.
Obvious from the previous remark and the general properties of parabolas (see Parabola.html ).

Remarks
[1] The discussion here is an alternative to the one in Artzt_Generation.html , where the use of projective (trilinear) coordinates gives an apparently simpler proof. Some more computations would be needed there to locate the directrix and the focus of the parabola, which here are determined explicitly, as is also the tangency point O.
[2] The properties proved here could be considered as a sequel or complement to the ones studied in ParabolaChords.html and ParabolaSkew.html .

Artzt.html
Artzt2.html
ArtztCanonical.html
Artzt_Generation.html
Parabola.html
ParabolaChords.html