Question

In $\Delta ABC$ with fixed length of $BC$, the internal bisector of angle $C$ meets the side $AB$ at $D$ and the circumcircle at $E$. The maximum value of $CD\times DE$ is $\frac{c^2}{k}$, then value of $k$ is

Answer 1

$CD.DE=AD.DB$
But $\frac{AD}{BD}=\frac{b}{a}$
$\Rightarrow AD=\frac{bc}{a+b}$ and $DB=\frac{ac}{a+b}$
$\therefore CD.DE=\frac{ab}{(a+b{)}^2}.c^2$

But $c^2$ is given, so we have to find the maximum value of
$\frac{ab}{(a+b{)}^2}$.

Now $\frac{ab}{(a+b{)}^2}=\frac{1}{\frac{a}{b}+\frac{b}{a}+2}$
But $\frac{a}{b}+\frac{b}{a}\ge 2$
then $\frac{a}{b}+\frac{b}{a}+2\ge 4$

$\therefore \frac{ab}{(a+b{)}^2}\le \frac{1}{4}$
Then max value of $CD.DE$ is $\frac{c^2}{4}$