TITLE: Given two parallel sides in a conic, the centre pass through line joining the midpoints.
QUESTION [3 upvotes]: Let $ABCDEF$ be a hexagon with opposite sides parallel.
Prove that the three lines joining the midpoints of
opposites sides are concurrent.

My progress: I could show that $ABCDEF$ lies in a conic. This is because consider $AB\cap ED, FE\cap BC,AF\cap CD.$ They are collinear. Hence by converse of pascal's theorem. We
get that $ABCDEF$ lies in a conic.
Now by diagram, it looks like the concurrency point is the centre of the conic. So what I conjecture is

Given two parallel sides in a conic, the centre pass through line joining the midpoints.

I did try to prove it. Note that this was true for circles. If we can somehow show that there exist a tomography which maps this conic to circle satisfying the properties we will be done. But we have midpoint, which is a non projective condition.

REPLY [1 votes]: Here is an approach using vectors. Say $M_1M_4$ and $M_2M_5$ intersect at point $O$. Now say, position vectors of $A, B, C, D, E$ and $F$ with respect to $O$ are $\mathbb{a}, \mathbb{b}, \mathbb{c}, \mathbb{d}, \mathbb{e}$ and $\mathbb{f}$ respectively.
As $AB$ and $DE$ are parallel,
$(\mathbb{a} - \mathbb{b}) \times (\mathbb{d} - \mathbb{e}) = 0 \tag1 $
$OM_1 = \frac{1}{2} (\mathbb{a} + \mathbb{b}) \ $,
$ \ OM_4 = \frac{1}{2} (\mathbb{d} + \mathbb{e})$
Given $M_1, O$ and $M_4$ are collinear,  $\vec {OM_1} \times \vec {OM_4} = 0$
$(\mathbb{a} + \mathbb{b}) \times (\mathbb{d} + \mathbb{e}) = 0 \tag2 $
Similarly,
$(\mathbb{b} - \mathbb{c}) \times (\mathbb{e} - \mathbb{f}) = 0 \tag3 $
$(\mathbb{b} + \mathbb{c}) \times (\mathbb{e} + \mathbb{f}) = 0 \tag4 $
$(\mathbb{a} - \mathbb{f}) \times (\mathbb{c} - \mathbb{d}) = 0 \tag5 $
So the problem statement reduces to showing $M_3, O$ and $M_6$ are collinear. In other words, we need to show that $\vec {OM_3} \times \vec{OM_6} = 0$
i.e. show that, $(\mathbb{a} + \mathbb{f}) \times (\mathbb{c} + \mathbb{d}) = 0$
Using $(5), \mathbb{a} \times \mathbb{c} + \mathbb{f} \times \mathbb{d} = \mathbb{a} \times \mathbb{d} + \mathbb{f} \times \mathbb{c}$
So, $(\mathbb{a} + \mathbb{f}) \times (\mathbb{c} + \mathbb{d}) = 2 (\mathbb{a} \times \mathbb{d} + \mathbb{f} \times \mathbb{c})$
Now expand and add $(1)$ and $(2)$ to get, $\mathbb{b} \times \mathbb{e} = \mathbb{d} \times \mathbb{a}$
and expand and add $(3)$ and $(4)$ to get, $\mathbb{b} \times \mathbb{e} = \mathbb{f} \times \mathbb{c}$
$ \implies \mathbb{f} \times \mathbb{c} = \mathbb{d} \times \mathbb{a}$
$\therefore (\mathbb{a} + \mathbb{f}) \times (\mathbb{c} + \mathbb{d}) = 2 (\mathbb{a} \times \mathbb{d} + \mathbb{f} \times \mathbb{c})$
$= 2 (\mathbb{a} \times \mathbb{d} + \mathbb{d} \times \mathbb{a}) = 0$
{"set_name": "stack_exchange", "score": 3, "question_id": 4233213}