# Given $3$ lines, prove that $3$ points cannot be collinear

Given $$3$$ distinct lines, $$A_1A_2$$, $$A_3A_4$$, and $$A_5A_6$$, which are concurrent at point $$P$$, prove that the lines $$A_2A_5$$ and $$A_1A_6$$ cannot both pass through point $$A_4$$.

When I drew out this diagram, it was quite evident to me that this statement is true, however, I have no idea how to actually formally prove this.

Any help would be greatly appreciated!

NOTE: All $$7$$ of these points are distinct

## Update

I'm not sure if I should make a new thread for this, so I'm just putting it here.

I thought of an interesting generalization to the problem, which I'm not too sure how to prove:

Given $$n$$ distinct lines, $$A_1A_2, A_3A_4, \cdots A_{2n-1}A_{2n}$$ which are all concurrent at point $$P$$, prove/disprove that the following $$n-1$$ lines can pass through $$A_{n+1}$$: $$A_1A_{2n}, A_2A_{2n-1}, A_3A_{2n-2} \cdots A_{n-1}A_{n+2}.$$

How would one go about solving this now?

• Actually, now that I draw the diagram, I don't see why this isn't possible. Take $A_1 = (1,0)$, $A_2 = (0,-2)$, $A_3 = (1,1)$, $A_4 = (0,0)$, $A_5 = (0,1)$, $A_6 = (-2,0)$, and $P = (2,2)$. Mar 17, 2021 at 19:30