[alogo] Parabola defined through similar triangles

[1] Start with a well known property of the tangents to a circle (c) from a point O. Draw from a point A of the circle orthogonals {AB0, AC0} to the tangents {OB,OC} and OD to line CB joining the contact points of the tangents. Then triangles ABC, AB0D and ADC0 are similar and AD2 = AB0*AC0.

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

[2] Generate similar triangles to the previous three by taking points {B2,C1} on line BC symmetric with respect to D and drawing on these sides triangles similar to ABC. Their other vertices {C2,B1} lie on lines {OC,OB} (see Similarly_Rotating.html ).
[3] Set AB/AC=k. Then AB1/AC2 = (AB1/AC1)/(AC2/AC1) = (AB1/AC1)/(AC2/AB2). Because of the similarity of triangles the last two ratios are respectively k and 1/k. Thus, AB1/AC2=k2.
[4] This implies that the right angled triangles {AB0B1, AC0C2} are similar and lines B1C2 cut off on the fixed lines {OB,OC} segments {B0B1,C0C2} in a fixed ratio: B0B1/C0C2 = k2.
[5] By the generalization of Thales theorem (see Thales_General.html ) lines B1C2 envelope a parabola which is tangent to lines {OB,OC,B0C0}.
[6] All circumcircles of triangles OB1C2 build a circle-bundle of intersecting type. One common point of all these circles is O the other is the focus F of the parabola.
[7] Line AD is also tangent to the parabola. This is seen by letting B2 take the position on BC such that B2C2 becomes parallel to B0D, B1C1 becoming then parallel to DC0. Then B1C2 coincides with AD.
[8] When B1C2 passes through A, measuring the angles there proves B1C2 to be parallel to BC and orthogonal to AD. Thus from A we have two tangents to the parabola intersecting there orthogonally and its polar GH passes through F. Hence A is on the directrix of the parabola. Besides, from general properties of the parabola, AF is orthogonal to the polar line GH of A, hence later passes through O, since AO is a diameter of the circumcircle of AOB.
[9] The parabola is easily constructible as a conic tangent to the five lines {OB,OC,A0B0,AD,AH}, later being orthogonal to AD.
[10] There are also two other, easy to construct positions of B1C2 (hence tangents to the parabola). These result by considering two special locations for points {B2,C1}. One when B1 coincides with B and the other when B2 coincides with B.
[11] Line OF is harmonic conjugate of OA with respect to lines {OB,OC}. This follows from O being on the polar of A (see [8]). Thus, by the reciprocity of polars, the polar of O passes also from A. This implies that {OB,OC,OA,OF} is a harmonic bundle.
[12] A consequence of the previous is that the tangent to circle (c) at A passes also through the intersection point I of BC and OF.
[13] Note that, according to general properties of triangles with sides tangent to a parabola, the tangent of the parabola at its vertex coincides with the Simson line of such a triangle with respect to the focus. Thus, this line in this case can be constructed by projecting F on line OB, OC.
[14] According to [11] F could be constructed by intersecting the circumcircle of OA0B0 with the harmonic conjugate line of OA with respect to {OB,OC}. Then project it to its Simson line ([13]) to obtain the vertex of the parabola. Proceed then to the parabola construction from its focus and its vertex.

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