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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0triangle similarity. FD/DR=CF/CE if fd>dr, cf is bound to be >ce

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry @GChamon what is DR?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait a sec... you still need to prove the similarity. it doesn't say anything else?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What kind of similarity did you use?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0cuz u get same angle, 90º, same side dc, but u still need another angle/side to prove the similarity

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sorry 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)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in a nutshell, the length of CF is dependent on length of FD, where the longer CF is the longer FD is.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that also applies to the other triangle

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's kind of difficult to understand for me. But let me think about it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is what I have until now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw: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'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0about 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand your approach, but could you express it in a more formal way.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I'm trying to do it but I can't figure out how to get from c>c' to a>a'

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I set up this equation \[a^2c^2=a'^{2}c'^{2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0If 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.

Callisto
 4 years ago
Best ResponseYou've already chosen the best response.0Not sure if it is correct but hope you understand it
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