$\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$ If bisector of angle C of an acute triangle ABC cuts the side AB in D and the circumcircle of triangle ABC in E,then $\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$.Then what is value of k?
Since angle bisector of C,bisects the arc BC.And circumcenter O will fall on line CE.And $\angle AOB=2\angle C$.These concepts i know but i caanot relate them to get answer.maybe i am lacking somewhere.Any suggestions or guidance is welcome.


$\frac{CE}{DE}=\frac{(a+b)^2}{kc^2}$. What is value of $k$?


Note that
\begin{align}
\angle EAB&=\angle EBA=\tfrac\gamma2
\end{align}
Assuming that
$|AB|=c$, $|BC|=a$, $|CA|=b$,
$\angle BAC=\alpha$,
$\angle ABA=\beta$,
$\angle ACB=\gamma$,
we can find $|CD|$ and $|CE|$ from the area relations:
\begin{align}
S_{\triangle ABC}&=
S_{\triangle ADC}+
S_{\triangle BDC}
\\
S_{CAEB}&=
S_{\triangle ABC}+S_{\triangle AEB}
=
S_{\triangle ACE}+S_{\triangle BCE}
\\
S_{\triangle AEB}&=
\tfrac14 c^2\tan\tfrac\gamma2
\end{align}
which gives
\begin{align}
|CD|
&=
\frac{ab\sin\gamma}{(a+b)\sin\tfrac\gamma2}
\\
|CE|
&=
\frac{ab\sin\gamma+\tfrac12 c^2\tan\tfrac\gamma2}{(a+b)\sin\tfrac\gamma2}
\\
|DE|&=|CE|-|CD|
=
\frac{\tfrac12 c^2\tan\tfrac\gamma2}{(a+b)\sin\tfrac\gamma2}
\\
\frac{|CE|}{|DE|}
&=
\frac{2ab\sin\gamma}{c^2\tan\tfrac\gamma2}+1
=
\frac{2ab(\cos\gamma+1)}{c^2}+1
\\
&=
\frac{a^2+b^2-c^2+2ab +c^2}{c^2}
=
\frac{(a+b)^2}{c^2}.
\end{align}
Hence $k=1$.
Edit:
Note that $|CE|$ is a common side of $\triangle BCE$ and $\triangle ACE$:
\begin{align}
S_{BCE}+S_{ACE}&=S_{ABC}+S_{AEB}
\\  
\tfrac12|CE|a\sin\tfrac\gamma2
+
\tfrac12|CE|b\sin\tfrac\gamma2
&=
\tfrac12ab\sin\gamma
+
\tfrac14c^2\tan\tfrac\gamma2
\\
|CE|\sin\tfrac\gamma2(a+b)
&=
ab\sin\gamma
+
\tfrac12c^2\tan\tfrac\gamma2
\\
|CE|&=
\frac{ab\sin\gamma
+
\tfrac12c^2\tan\tfrac\gamma2}{\sin\tfrac\gamma2(a+b)}.
\end{align}
Edit 2:
In $\triangle AEB$ we have $\angle EAB=\angle EBA=\tfrac\gamma2$,
thus it's isosceles and $|AE|=|BE|$.
Its area $S_{AEB}=\tfrac12 h |AB|$, where 
$h$ is the altitude, which splits 
isosceles $\triangle AEB$
into two congruent right triangles 
with the base $\tfrac12c$, so
$h=\tfrac12|AB|\tan\angle EAB=
\tfrac12c \tan\tfrac\gamma2$,
hence $S_{AEB}=\tfrac12 h c=\tfrac14c^2 \tan\tfrac\gamma2$.
