Proof for parallel lines

Marcus-Tullius

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Jan 13, 2019
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Hello

I have to proof that
GH || FE when alpha = 45 degrees

Number 3.jpg

I don´t have any idea of an approach. So please help.
Thanks in advance.
 
Hello

I have to proof that
GH || FE when alpha = 45 degrees

View attachment 10883

I don´t have any idea of an approach. So please help.
Thanks in advance.

I constructed this on GeoGebra and confirmed that it is true with no additional conditions needed; for example, E does not have to be where it is shown.

I don't yet have a proof, but one idea is to construct the tangent to the circle at A and find an angle that is congruent to the inscribed angle AGH. You then want to show that this angle is also congruent to AFE.

(I may be reading some letters wrong, because the picture is a little small.)
 
I'm sorry. This picture should be better.

I also have the information that E can be anywhere on BC and F on CD. But E and F cannot be at the same place with another point.

I will try to solve it with the tangent at A, but I would appreciate any other help.
 

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I'm sorry. This picture should be better.
I also have the information that E can be anywhere on BC and F on CD. But E and F cannot be at the same place with another point.
I will try to solve it with the tangent at A, but I would appreciate any other help.
parallel.jpg angle
This is a much better image for there is no implied additional information.
Note that if \(\displaystyle m(\angle HAG)=\dfrac{\pi}{4}\) then \(\displaystyle m(arc {HCG})=\dfrac{\pi}{2}\) that is a quarter of the circle.
So I must agree with Prof Khan there has got to be more given.
Here is a thought: the OP showed the diagram on grid. Are we to assume \(\displaystyle ABCD\) is a rectangle(a square)???
 
Last edited:
I've been looking at the problem from time to time, and I finally found a chain of congruent angles that will work. (I'd been just looking, and not writing enough down.)

If you've seen that angle HGA is congruent to the angle between AH and the tangent, you can show that this is congruent to BCG = complement of DAG = DFA = EFA, which is what you need to show that GH and FE are parallel.

I leave you to fill in the gaps.
 
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