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

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.

• As far as the original problem, it can proved using vectors too. Are you open to that? Commented Aug 26, 2021 at 3:42
• @MathLover aa yes! Commented Aug 26, 2021 at 3:50
• 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.
– Pope
Commented Aug 26, 2021 at 6:49
• 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. Commented Aug 26, 2021 at 7:07
• woah i see, thanks! Commented Aug 26, 2021 at 7:12

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, $$\mathbb{b} \times \mathbb{e} = \mathbb{f} \times \mathbb{c}$$

$$\implies \mathbb{f} \times \mathbb{c} = \mathbb{d} \times \mathbb{a}$$

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

• 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 Commented Aug 26, 2021 at 4:54
• I will edit with more details. In fact I missed to add one equation. Commented Aug 26, 2021 at 4:56
• 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! Commented Aug 26, 2021 at 5:41