Question

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

Answer 1

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

Answer 2

sorry @GChamon what is DR?

Answer 3

sry! typo. DE

Answer 4

wait a sec... you still need to prove the similarity. it doesn't say anything else?

Answer 5

What kind of similarity did you use?

Answer 6

cuz u get same angle, 90º, same side dc, but u still need another angle/side to prove the similarity

Answer 7

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)

Answer 8

in a nutshell, the length of CF is dependent on length of FD, where the longer CF is the longer FD is.

Answer 9

that also applies to the other triangle

Answer 10

It's kind of difficult to understand for me. But let me think about it.

Answer 11

This is what I have until now

Answer 12

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

Answer 13

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?

Answer 14

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'