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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?

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    $\begingroup$ 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 $P = (2,2)$. $\endgroup$ Mar 17, 2021 at 19:30

2 Answers 2

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I think it's possible. Let me know if I'm missing something. A counterexample

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Here's a counter-example to your "answer":

enter image description here

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