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.

enter image description here

  • 1
    $\begingroup$ As far as the original problem, it can proved using vectors too. Are you open to that? $\endgroup$
    – Math Lover
    Commented Aug 26, 2021 at 3:42
  • 1
    $\begingroup$ @MathLover aa yes! $\endgroup$ Commented Aug 26, 2021 at 3:50
  • 1
    $\begingroup$ By sides of a conic, I see you must mean chords. The line joining the center to the midpoint of a chord is a diameter, and the chord is in the corresponding ordinate direction. This diameter must bisect all parallel chords. I am not actually familiar with Pascal's proof, but it must surely depend on fundamental properties like this one, known and proved long before his time. $\endgroup$
    – Pope
    Commented Aug 26, 2021 at 6:49
  • 1
    $\begingroup$ The line joining the midpoint of a chord with the center of a conic, intersects all the other parallel chords at their midpoint. This is a basic property of conics. $\endgroup$ Commented Aug 26, 2021 at 7:07
  • $\begingroup$ woah i see, thanks! $\endgroup$ Commented Aug 26, 2021 at 7:12

1 Answer 1


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 $


$(\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}$

$\therefore (\mathbb{a} + \mathbb{f}) \times (\mathbb{c} + \mathbb{d}) = 2 (\mathbb{a} \times \mathbb{d} + \mathbb{f} \times \mathbb{c})$
$= 2 (\mathbb{a} \times \mathbb{d} + \mathbb{d} \times \mathbb{a}) = 0$

  • $\begingroup$ Okie well I got by expanding, (axd)+(bxe)=0, (bxe)+(cxf)=0 or (fxc)+(exb)=0. So we get (axd)+(fxc)=0, so we just want (fxd)+(axc)=0 $\endgroup$ Commented Aug 26, 2021 at 4:54
  • 1
    $\begingroup$ I will edit with more details. In fact I missed to add one equation. $\endgroup$
    – Math Lover
    Commented Aug 26, 2021 at 4:56
  • 1
    $\begingroup$ Yes thankyou I got it! The last two step wasn't necessary if? Since fxc=-axd, and we already got axc+fxd=0, so sum is 0. Thanks a lot for help! $\endgroup$ Commented Aug 26, 2021 at 5:41

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .