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

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

that also applies to the other triangle

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

This is what I have until now

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 \[a^2-c^2=a'^{2}-c'^{2}\]

Not sure if it is correct but hope you understand it