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I would appreciate some help with this proof: Let \[CD\perp AB, FD > DE\], prove that \[CF>CE\]

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triangle similarity. FD/DR=CF/CE if fd>dr, cf is bound to be >ce
sorry @GChamon what is DR?

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Other answers:

sry! typo. DE
wait a sec... you still need to prove the similarity. it doesn't say anything else?
What kind of similarity did you use?
cuz u get same angle, 90º, same side dc, but u still need another angle/side to prove the similarity
sorry dude, no need for similarity, it is basic triangle properties. they (triangle CFD and CDE) both share same side CD and same angle 90º. you have then that a^2=b^2+c^2 lets say they both share same 'c' side. b(CFD)>b(CDE ,<=> a(CFD)>a(CDE)
in a nutshell, the length of CF is dependent on length of FD, where the longer CF is the longer FD is.
that also applies to the other triangle
It's kind of difficult to understand for me. But let me think about it.
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This is what I have until now
i find your steps a little reduntant. try to observe the fact that they both share the same side and have 90º angles. you can relate each other using a^2=b^2+c^2
I'm going to do it the way you're suggesting. But I just wanted to know if the work I did makes sense and indeed is correct?
|dw:1333245486002:dw| b^2+c^2=a^2 b^2+c'^2=a'^2 if b is the same in both equations, the longer the c, the longer the a. if c>c', a>a'
about your resolution, i couldn't really tell if it was going to the right direction, could work, but i am not sure. anyway mathematically it is correct
I understand your approach, but could you express it in a more formal way.
I'm trying to do it but I can't figure out how to get from c>c' to a>a'
I set up this equation \[a^2-c^2=a'^{2}-c'^{2}\]
If two triangles have two angles equal to two angles respectively, and one side equal to one side, which may be either the sides between the equal angles or the sides opposite one of them, then the remaining sides will equal the remaining sides (those that are opposite the equal angles), and the remaining angle will equal the remaining angle.Let ABC, DEF be two triangles in which angles B and C are equal respectively to angles E and F; let the adjoining side BC be equal to the adjoining side EF; then AB will equal DE, AC will equal DF, and angle A will equal angle D. The proof is not really difficult but it involves two cases; this one for when the adjoining sides BC, EF are equal, and another for when the sides opposite equal angles are equal: sides AB, DE; hence the proof is lengthy. Each case is proved by the indirect method, and rests ultimately on S.A.S.
Not sure if it is correct but hope you understand it
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