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The Problem enter image description here

So, I was able to solve this problem, but what I'm curious of is once we solve the variables can I extend point F to line BC making Triangle ABX? I imagine an extension of point F would have to lie on BC at some point and intersect it. And because we're told that the x variables are part of a tri-section, each angle would have to be the same measure.

When I first saw the problem I wanted to extend it and try and prove triangle ABX might be isosceles, but my instructor informed me that it was not possible, which I can see why now. But what about AFTER the variables are discovered, shouldn't that allow us to see where line AF would extend to?

Or, would I need this shape to lie on a coordinate plane to be able to determine all this? Thanks in advance!

Edit to clarify: I want to determine where the extension of segment AF would intersect segment BC making the Triangle ABX (where X is the new point extended from segment AF). Is it even possible to do so?

Joffan
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Insalubrious Sophist
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  • Are you wanting to extend segment AF until it intersects segment BC? Your first paragraph is not clear ("an extension of point F" makes no sense). – scott Aug 10 '17 at 21:42
  • Sorry if I was vague, but yes, I want to extend segment AF until it intersects segment BC. – Insalubrious Sophist Aug 10 '17 at 21:44
  • It's not actually known whether $F$ is inside the quadrilateral, from the definitions. $x$ and $y$ are not explicitly solved for. – Joffan Aug 10 '17 at 22:00
  • Hm, do you mean because everything isn't coplanar we can't know? – Insalubrious Sophist Aug 10 '17 at 22:04
  • No, even assuming coplanar, because we don't know the lengths of the sides. – Joffan Aug 10 '17 at 22:06
  • I see your point, thank you for replying and helping, Joffan! – Insalubrious Sophist Aug 10 '17 at 22:08
  • I’m not sure what you mean by “discovering” or “solving for” the variables here. The point of the original problem is that even though the values of angles $x$ and $y$ themselves cannot be determined, their sum is constant. – amd Aug 10 '17 at 23:55

2 Answers2

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If $AF\cap BC=\{X\}$ then $X$ can be placed not on segment $BC$.

Also, $x$ changes and you can not prove, what you want.

Michael Rozenberg
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By changing the lengths $AB$ and $CD$ (or equivalently $BC$ and $DA$), we can manipulate the intersection of $AF$ with $BC$ or $CD$ within wide limits. Quite apart from the difficulty of not knowing the individual angle values of $x$ or $y$, only the sum.

It's reasonable to assume the figure is on a plane, though. That way the intersections are somewhat predictable.

Joffan
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  • I'm not sure why the sum of x + y vs the values of x and y individually never crossed my mind. A subtle, but interesting point that blew over my head! Well, maybe not so subtle, considering you're substituting for (2x+2y) in 180 - (2x+2y), hah. – Insalubrious Sophist Aug 10 '17 at 22:16