## Parabola trapezium

Consider a parabola (c) and the triangle formed by two tangents AB, AC and their chord of contacts BC. Line DE joining the middles of AB, AC is also tangent to (c) at its middle K. The diagonals CE, BD of the trapezium BCDE intersect (c) at points J, I, which projected parallel to the axis on BC define F, H dividing BC in three equal parts. Also the tangents at J, I are parallel to the other diagonals correspondingly and GI = 2*ID.

To prove it use a canonical parabola (c) defined analogously to the previous but starting with an equilateral triangle ABC. Prove the theorem for that case and use an affinity to transfer the property to the general case. To start with, let I be the intersection point of BD with the parabola. Consider the circumcircle of the triangle of tangents K'DC', which passes also through the focus G. By the elementary properties of the parabola K' is the middle of GH', H' being the projection of I on the directrix line (e). It follows that KK' = K'I' = I'D and the tangent IC' is parallel to EC. Thus all statements made above are easily verified in this special configuration.
To handle the general case consider the affinity F mapping the equilateral to the general triangle. Since F preserves ratios and parallelity, it maps the whole configuration to the corresponding of the general case, thus transfering the relations verbatim.

By the way, notice that this special parabola (its reflexion on the horizontal axis DE) is represented in orthogonal axes in which |AG|=2 by the function y = x2.