<s>[INST] 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 $CD=\frac{2ab}{a+b}\cos\frac{C}{2}$
 [/INST] Complete the cyclic quadrilateral $ACBE$ and note,
$$\angle ABE=\angle ACE=\angle BCE=\angle BAE=\frac{\angle C}{2}\implies AE=BE.$$
Using the Angle Bisector Theorem,
$$\frac{a}{b}=\frac{BD}{AD}\implies \frac{a+b}{b}=\frac{c}{AD}\implies AD=\frac{bc}{a+b}$$
Since $\triangle ADC\sim\triangle EDB,$
$$\frac{b}{AD}=\frac{BE}{DE}\implies BE=DE\cdot\frac{a+b}{c} $$
Applying Ptolemy's Theorem,
$$\begin{align*}AE\cdot a+BE\cdot b&=CE\cdot c\\
BE\cdot(a+b)&=CE\cdot c\\
DE\cdot\frac{a+b}{c}\cdot (a+b)&=CE\cdot c \\
\therefore\; \frac{CE}{DE}&=\left(\frac{a+b}{c}\right)^2
\end{align*}$$
</s>