ParabolaBySimilarTriangles

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.

[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.

Parabola defined through similar triangles

See Also