Equidistant Parallel Lines

Property discovered by Ada while researching for new paper folding patterns.

There are 5 equidistant parallel lines A0B0, A1B1, A2B2, A3B3 and A4B4.

The line CD intersects A2B2, A3B3 and A4B4 in F2, F3 and F4.
The line CE intersects A1B1, A2B2 and A3B3 in G1, G2 and G3.

F2G1 intersects A0B0 in H0.
F3G2 intersects A1B1 in H1.
F4G3 intersects A2B2 in H2.

Then H0, H1 and H2 are collinear.

Proof:

We first notice that:
$ \overrightarrow{F_3F_2}+\overrightarrow{F_3F_4}=0 $
$ \overrightarrow{G_1G_2}+\overrightarrow{G_3G_2}=0 $

Because this are alternative routes from $ F_3 $ to $ G_2 $ we have:
$ \overrightarrow{F_3G_2}=\overrightarrow{F_3F_2}+\overrightarrow{F_2G_1}+\overrightarrow{G_1G_2} $
$ \overrightarrow{F_3G_2}=\overrightarrow{F_3F_4}+\overrightarrow{F_4G_3}+\overrightarrow{G_3G_2} $

We add the last two equalities and get:
$ 2\overrightarrow{F_3G_2} = (\overrightarrow{F_3F_2}+\overrightarrow{F_3F_4}) + (\overrightarrow{F_2G_1}+\overrightarrow{F_4G_3}) + (\overrightarrow{G_1G_2}+\overrightarrow{G_3G_2}) $

But the first and third from the right side are 0:
$ 2\overrightarrow{F_3G_2}=\overrightarrow{F_2G_1}+\overrightarrow{F_4G_3} $

We multiply the last equality by 2 and get:
$ 4\overrightarrow{F_3G_2}=2\overrightarrow{F_2G_1}+2\overrightarrow{F_4G_3} $

We also notice that:
$ 2\overrightarrow{F_2G_1} = \overrightarrow{F_2H_0} $
$ 2\overrightarrow{F_3G_2} = \overrightarrow{F_3H_1} $
$ 2\overrightarrow{F_4G_3} = \overrightarrow{F_4H_2} $

So if we do the replacements we get:
$ 2\overrightarrow{F_3H_1} = \overrightarrow{F_2H_0}+\overrightarrow{F_4H_2} $ (1)

Because this are alternative routes from $ F_3 $ to $ H_1 $ we have:
$ \overrightarrow{F_3H_1} = \overrightarrow{F_3F_2}+\overrightarrow{F_2H_0}+\overrightarrow{H_0H_1} $
$ \overrightarrow{F_3H_1} = \overrightarrow{F_3F_4}+\overrightarrow{F_4H_2}+\overrightarrow{H_2H_1} $

We add the last two equalities and get:
$ 2\overrightarrow{F_3H_1} = (\overrightarrow{F_3F_2}+\overrightarrow{F_3F_4} ) + (\overrightarrow{F_2H_0} + \overrightarrow{F_4H_2}) + (\overrightarrow{H_0H_1}+\overrightarrow{H_2H_1}) $

but the first from the right is zero
$ 2\overrightarrow{F_3H_1} = (\overrightarrow{F_2H_0} + \overrightarrow{F_4H_2}) + (\overrightarrow{H_0H_1}+\overrightarrow{H_2H_1}) $

because of (1) we have:
$ \overrightarrow{H_0H_1}+\overrightarrow{H_2H_1} = 0 $

so $ H_0, H_1, H_2 $ are collinear

Note:

This property leads to the following recursive pattern:

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