Similar Triangles Proof (Geometry)

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gullin15

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A friend of mine stumped me with this math problem, I can't seem to think I'm forgetting something...
After about an hour I think I solved it but I'm not sure so I'm going to need some help! Thank you.
mathproblem.jpg
Edit:fixed

I believe that you have to prove DE and AB parallel and then use the vertical angles from DCE and CAB as well as an alternate exterior angle to prove them similar by AA~. I'm just not too sure on how to prove DI || AB.

Edit 2: Just figured it out. Instead of finding any lines parallel I used proportions after AA~ and then used SAS to prove ABC ~ DCE.
 
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gullin15 said:
A friend of mine stumped me with this math problem, I can't seem to think I'm forgetting something...
After about an hour I think I solved it but I'm not sure so I'm going to need some help! Thank you.
View attachment 3800.
Click to expand...
Now I have not done the work. But I think the is a matter of chasing smaller similar triangles and congruent angle.
I also think that you need to label two more points: \(\displaystyle \left\{ J \right\} = \overline {AB} \cap \overline {CF} \;\& \;\left\{ K \right\} = \overline {DE} \cap \overline {HC} \)

You can observe things like \(\displaystyle m\left( {\angle DCA} \right) = m\left( {\angle CAB} \right) + m\left( {\angle CBA} \right)\) and \(\displaystyle m\left( {\angle CJA} \right) = m\left( {\angle FJB} \right)\).
 
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Like pka, I have not done the work, but I am wondering whether anything else is given, such as hcf is a straight line.
 
Straight lines can be assumed from the diagram, so yes. But other than that only that is given. After thinking a bit more, I'm not sure how to prove DB parallel to both HG and EF, because from there it's easy to figure out.
 
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