Can you prove or disprove the following claim:
Construct a hexagon circumscribed around a conic section. Intersection points of its non-principal diagonals lie on a new conic section.
GeoGebra applet that demonstrates this claim can be found here.
A conic inside a hexagon
conic sectionseuclidean-geometrygeometrypolygonsprojective-geometry
Related Solutions
$\newcommand{real}{\operatorname{Re}}$Set up a coordinate system where the origin is the pentagon centre; without loss of generality take the pentagon's inradius as $1$ and the tangent line as $x=-1$. Then $a=2\tan\frac\pi5$, the pentagon circumradius $R=\sec\frac\pi5$ and $K=5\tan\frac\pi5$.
The argument of $A$ may be any angle $\theta$, but then $d_1=1+\real(Re^{i\theta})=1+R\real(e^{i\theta})$. Thus $$d_1+d_2+d_3+d_4+d_5=5+R\real\left(\sum_{k=0}^4e^{i(\theta+2k\pi/5)}\right)$$ The terms in the sum of exponentials are the roots of $z^5=e^{5i\theta}$, so by Viète's relations that sum is the negative of the $z^4$ coefficient, which is zero. Therefore $d_1+d_2+d_3+d_4+d_5=5$ and the equation to prove reduces to $5a=2K$, which is easily seen to be true.
The short answer is that the three circles are coaxal, and therefore their centers are collinear. The circles are coaxal because of the following lemma (screen shot from Johnson, Modern Geometry aka Advanced Euclidean Geometry, pg 92):
Proof later, but you might take a look at another question and answers.
All of this is related to Poncelet's Porism, one general version of which is (excerpted from Hraskó, Poncelet's theorem, well worth reading in full):
Poncelet's General Theorem: Let $e$ be a circle of a non-intersecting pencil and let $a_1,a_2,\ldots,a_n$ be (not necessarily different) oriented circles in the interior of $e$ that belong to the same pencil. Starting at an arbitrary point $A_0$ of the circle $e$, the points $A_1,A_2,\ldots,A_n$ are constructed on the same circle, such that the lines $A_0A_1, A_1A_2, \ldots, A_{n-1}A_n$ touch the circles $a_1,a_2,\ldots, a_n$, respectively, in the appropriate direction. It may happen that at the end of the construction, we get back to the starting point, that is, $A_n=A_0$. The theorem states that in that case, we will always get back to the starting point in the $n$-th step, whichever point of $e$ we start from. We do not even need to take care to draw the tangents to the circles in a fixed order.
Here's a page from Poncelet's Traité des propriétés projectives des figures, 1827 that illustrates the lemma.
But we don't need the general theorem, other than to observe that a Poncelet Porism will involve (or generate) circles in a coaxal system.
By Poncelet's porism, you can move the pentagon in the OP continuously with its vertices on the circumcircle and sides touching the incircle. Along for the ride will go triangles like $A_1A_2A_3$ with sides $A_2A_1,A_2A_3$ tangent to the incircle and line $d_2$ which must be tangent to the diagonals incircle (it's shape will vary). By the lemma, the diagonals incircle must be coaxal with the other two circles.
The proof of the lemma is excerpted here (Johnson, pg 93):
Best Answer
This is a consequence of Pascal and Brianchon's Theorems.
The intersections of the non-principal diagonals can also be seen as the intersections of the triangles $\triangle{DBF}$ and $\triangle{CAE}$. By Brianchon's theorem, the principal diagonals $EB,FC,AD$ are concurrent at a point $X$. Thus the two triangles are perspective. A converse of Pascal's theorem says that the points of intersection of two perspective triangles lie on a common conic.
Details and more precise statements can be found in Hatton's Projective Geometry, pg 189. There you can find Pascal's theorem, its converse, Brianchon's theorem and proofs.