Draw 7 lines on the plane in an arbitrary manner. Prove that for any such configuration, 2 of the those 7 lines form an angle less than 26◦ I have been working on this question for a while now and I think that this is one of the many applications of The Pigeonhole Principle. However, I don't seem to draw a conclusion.
So, I figured that the lines must intersect somehow to form a Heptagon and the sum of the exterior angles must be 360 degrees which would be distributed among the pairs of lines that will be formed. I also noticed that $\frac{180}{7}=25.714$ approximately which incentivized me to carry out this procedure, however, I don't see a continuation. Thanks!
Edit: It seems that the case of 2 lines being parallel is breaking the statement of the title, so I think it is safe to assume that we are talking about taking 7 lines arbitrarily where no 2 lines are parallel to each other.
 A: It was a bit tricky to apply the pigeonhole principle to this problem. I initially thought about dividing a $180^\circ$ protractor into seven equal sections which were the holes, and the counter-clockwise angles the seven lines made with the $x$-axis were the pigeons. Certainly, if two lines are in the same hole, then their angle is at most $180/7<26$. However, this is seven pigeons in seven holes, so we cannot conclude there are two cohabitating pigeons.
Here is what I found does work. Arbitrarily single out one of the lines, and call it $L$. If any of the other six lines intersect $L$ at an angle of at most $180/7$ degrees, then we are done. Otherwise, the measure of the counterclockwise angle each other line makes with $L$ is between $180\cdot \frac17$ and $180\cdot \frac67$. We can divide the interval $[180\cdot \frac17,180\cdot \frac67]$ into five intervals of length $180/7$. These are our five holes, and the angles with $L$ of the six other lines are the pigeons.
A: Big hint, with seven lines differing maximally in angle:

Because you are interested just in the relative angles, you can arbitrarily shift each line to go through the same point (the origin).
Do you think there is a way to make all the angles greater than $26^\circ$?
Well I suppose if you're allowed to construct seven non-intersecting parallel lines... well then sure.  But this happens statistically with measure $0$.
A: This is false.
Consider the case of three lines parallel to each other, and perpendicular to four other lines. All of the angles between them will be 90 degrees. Since 90 is greater than 26, this conjecture is false. QED.
You can also do this by drawing a regular hexagon out of lines, and then drawing an additional line between two of the opposite corners, producing four thirty degree angles and several 120 and 60 degree angles.
