Pasch's Theorem for Asymptotic Triangles

[ILoveJesseJam]

Theorem: If a straight line intersects one of the sides of the asymptotic triangle ABOmega but does not pass through a vertex (including omega) it will intersect exactly one of the other two sides.

Proof.

I have a few ideas for this proof. First, I think it will be three cases where the straight line intersects AB, Aomega and Bomega. I have a few questions. I have the other proof for when the straight line passes through one of the vertexes of triangle ABomega it will intersect the other side. With this, AC was drawn to be the perpendciular to Bomega through A. Do I need to construct this line again for this above proof? I was thinking I did not.. But, was not sure. I did do it when the line intersected AB in one version of this proof.. Making it two cases for that situation. I made it when AC fell outside of triangle ABomega and when it fell inside triangle ABomega. I'm not sure if this is correct... If I need the perpendicular line in all three cases of AB, Aomega, or B omega.. or none at all.

With the line intersecting Aomega in one case and Bomega in the other, I thought I could write one proof for it and say it will be the same for when it intersected the other side. I was thinking: Let l intersect Bomega and contain the interior point H of triangle ABomega. Construct AH. By the previous theorem, AH intersects Bomega at T. Consider triangle ABT. Since l intersects BC and point H on AC, it must intersect either A omega or C omega. Since C is a an element of B omega, it must then intersect A omega, the other side of B omega.
Am I on the right track? I've looked at this proof a long time.. And, I'm just somewhat lost. Thank you for reading and assisting me.

 

[jvicens]

Thank you for your post and for sharing your ideas for the proof. It seems like you have a good understanding of the problem and are on the right track. I would suggest completing the proof by considering all three cases individually, as you mentioned. This will ensure that your proof covers all possible scenarios.

In the case where the line intersects AB, you can use a similar approach as the one you used for the previous theorem. You can construct a perpendicular line from point A to the line l, and show that it must intersect either Aomega or Bomega. This will prove that the line intersects exactly one of the other two sides.

In the case where the line intersects Aomega, you can use the same reasoning as you did for the previous theorem. Construct a line from point B to the line l, and show that it must intersect either AB or Bomega. This will prove that the line intersects exactly one of the other two sides.

Finally, in the case where the line intersects Bomega, you can use the same approach as you did for the case where the line intersects Aomega. Just construct a line from point A to the line l, and show that it must intersect either AB or Aomega.

By completing all three cases, you will have proven that the line intersects exactly one of the other two sides in all scenarios. I hope this helps and clarifies your thoughts. Good luck with your proof!