If bisector of angle $C$ of $\triangle ABC$ meet $AB$ in $D$ and circumcircle in $E$ prove that $\frac{CE}{DE}=\frac{(a+b)^2}{c^2}$. If bisector of angle $C$ of $\triangle ABC$ meets $AB$ in point $D$ and the circumcircle in point $E$ then prove that $$\frac{CE}{DE}=\frac{(a+b)^2}{c^2}$$.
My Attempt
Using the fact that $$CD=\frac{2ab}{a+b}\cos\frac{C}{2}$$
and $$AD.DB=CD.DE$$
we get$$\left(\frac{bc}{a+b}\right)\left(\frac{ac}{a+b}\right)=\left(\frac{2ab}{a+b}\cos\frac{C}{2}\right)DE\Rightarrow DE=\frac{c^2}{2(a+b)\cos\frac{C}{2}}$$ and then I did lot of calculation to obtain $$CE=CD+DE=\frac{a+b}{\cos\frac{C}{2}}$$ to get the final ratio.
But I believe there would be a more generic way to do it perhaps a geometrical solution rather than relying upon the formula of angle bisector $CD=\frac{2ab}{a+b}\cos\frac{C}{2}$

Using formula for length of angle bisector, $ \ \displaystyle CD^2 = \frac{ab}{(a+b)^2} ((a+b)^2 - c^2)$
By Intersecting Chords theorem,
$ \displaystyle CD.DE = AD \cdot DB = (\frac{a}{a+b} \cdot c) \cdot (\frac{b}{a+b} \cdot c) = \frac{ab c^2}{(a+b)^2}$
$ \displaystyle \frac{CE}{DE} = 1 + \frac{CD^2}{ CD \cdot DE} = 1 + \frac{(a+b)^2}{abc^2} \cdot CD^2 = 1 + \frac{(a+b)^2 - c^2}{c^2} = \frac{(a+b)^2}{c^2}$
