Proving Ptolemy's inequality with vectors

[*Mahi*]

Try proving Ptolemy's inequality by vectors.
(I think this is useful for the inequality part $CD.EF+DE.CF\geq CE.DF$)

[*Mahi*]

Solution:

$\overrightarrow{FD}\times\overrightarrow{CE}=(\overrightarrow{FC}+\overrightarrow{CD})\times(\overrightarrow{CD}+\overrightarrow{DE})$
$=(\overrightarrow{FC}\times\overrightarrow{CD}+\overrightarrow{FC}\times\overrightarrow{DE}+\overrightarrow{CD}\times\overrightarrow{CD}+\overrightarrow{CD}\times\overrightarrow{DE})$
$=(\overrightarrow{FC}\times\overrightarrow{DE}+\overrightarrow{CD}(\overrightarrow{FC}+\overrightarrow{CD}+\overrightarrow{DE}))$
$=(\overrightarrow{FC}\times\overrightarrow{DE}+\overrightarrow{CD}\times\overrightarrow{EF})$
Now taking lengths on both sides,
$\left | FD \right |\left | CE \right |\leq\left | FC \right |\left | DE \right |+\left | CD \right |\left |EF \right |$
So,we are done.

[Moon]

Hmm...that's cool. You can also use complex numbers; you already know that, right? However, like always I would always recommend you to learn more euclidean geometry.

[*Mahi*]

Yeah I know that of course.......but I thought in many cases,vector and complex is the same thing we call by two names.........as in vectors $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ is all the same as $(\alpha - \beta) +(\beta -\gamma )=( \alpha - \gamma )$ ,isn't it?

[*Mahi*]

But I think I will be posting the 'complex' one within a little time..........