# Three properties of a parabola with three tangents

About ten days ago I discovered three beautiful properties of a parabola with three tangents drawn using the GeoGebra program. I would like to get proof of these properties and also know if any of them have been discovered before or not.

We have a parabola, and we chose three points from it, $$A$$, $$B$$, and $$C$$. We drew the tangents to the parabola in them, which enclose the triangle $$∆MNP$$, so the area of triangle ABC will be equal to twice the area of triangle $$∆MNP$$. In other words $$T=2S$$

Another beautiful property is achieved in the form: $$\frac{FA×FB×FC}{FM×FN×FP}=1$$

Also, the lines $$AN$$, $$BM$$, and $$CP$$ converge at one point

• The first property is known : see here. I didn't know the second property nor the third. Commented Nov 22, 2023 at 20:09
• A little remark : you don't need figure 1. Figure 2 is enough. Commented Nov 22, 2023 at 22:12
• A result that could be helpful : Lambert's theorem asserting that the circumcircle of triangle MNP passes through focus $F$. Commented Nov 22, 2023 at 22:43
• Another useful result : the fact that the orthocenter of triangle $MNP$ belongs to the directrix of the parabola can be found in this answer of mine here. Commented Nov 23, 2023 at 1:55
• Thank you very much, yes I know Lambert's theorem, but I did not know that the first property was previously discovered, but it is expected because it is an elementary property of the parabola Commented Nov 23, 2023 at 2:03

## 1 Answer

Let's prove the third property in the special case of a circle: as projective transformations preserve collinearity, the result holds for any conic section.

The proof follows from the converse of Ceva's theorem applied to triangle $$PMN$$: lines $$PC$$, $$AN$$, $$BM$$ concur if:

$${BN\over BP}\cdot{AP\over AM}\cdot{CM\over CN}=1.$$ But that is trivially true, because $$BN=CN,\quad BP=AP,\quad AM=CM.$$