[alogo] Thales Parabola

Given two intersecting lines GE, GH and a point F not lying on these lines. There is a parabola having the given lines as tangents and also having as tangents all the lines UW, determined by the intersection point U, W of the two lines with all the circles of the circle bundle of circles passing through G and F.
Point F is the focus of the parabola and its directrix can be determined easily through the given data (the two lines GE, GH and point F).

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]
[2_0] [2_1] [2_2] [2_3]

Start with the four lines {GE, GH, UW, VN}. The key to the proof is Miquel's theorem (see Miquel_Point.html ) and its consequences. By these the circumcircles of the four triangles {GWU, GNV, IWN, IUV} formed by these four lines pass all through point F. These four lines define also a parabola to which they are tangent and which has at F its focus.
From the properties of parabolas and the generalized Thales theorem (see Thales_General.html ) it follows that all lines PR resulting from arbitrary circles passing through {F,G} and their intersections with lines {GE, GH} are all tangent to the parabola.
To find the directrix consider the minimal circle of the circle-bundle through {G,F} which is the circle with diameter GF. Let {Q, X} be the intersections of this circle with lines {GE, GH}. Consider also line FE', which is orthogonal to line QX. The circle through {F, G, E'} intersects line GH at H'. Comparing the angles at E', Q and G we see that angle(FE'H) = angle(FE'Q) (taking into account that X is the middle of the chord GH'). Hence line FE' is the axis of the parabola, since E'H' and E'G are two tangents symmetric with respect to this line. From the aforementioned Miquel results we know that the orthocenter Y of the isosceles triangle E'QZ is on the directrix of the parabola. We see also easily that QYZF is a rhombus, hence Y is the symmetric of F with respec to QX.
The recipe of the directrix construction is thus:
1) Construct the circle on diameter FG.
2) Find the intersections {Q,X} of this circle with lines {GE, GH}.
3) Find the symmetric Y of F with respect to line QX.
4) The directrix is the parallel to QX from Y. The directrix passes through the orthocenter Y of the isosceles QE'Z. And GQYZ having its diagonals bisecting orthogonally is a rhombus.

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