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

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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