Question

In a $△ABC$ bisector of angle $C$ meets the side $AB$ at $D$ and circumcircle at $E$. The maximum value of $CD.DE$ is equal to

• A. $\frac{a^2}{4}$
• B. $\frac{b^2}{4}$
• C. $\frac{c^2}{4}$
• D. $\frac{{(a+b)}^2}{4}$

Answer 1

The correct option is C $\frac{c^2}{4}$

In a $△ABC$,
$\frac{AD}{BD}=\frac{AC}{BC}=\frac{b}{a}$ ...............$\left(1\right)$
and $CD.DE=AD.BD$ ..................$\left(2\right)$
Now, from eqn$\left(1\right)$
$\frac{AD}{b}=\frac{BD}{a}$
$=\frac{AD+BD}{b+a}$
$=\frac{AB}{a+b}$
$=\frac{c}{a+b}$
$\therefore AD=\frac{bc}{a+b},BD=\frac{ac}{a+b}$ $\left(3\right)$
From eqns$\left(2\right)$ and $\left(3\right)$, then
$CD.DE=\frac{ab{c}^2}{{(a+b)}^2}$
$\because$A.M$\ge$G.M
$\Rightarrow \frac{a+b}{2}\ge \sqrt{ab}$
Squaring both sides, we get
$\Rightarrow {\left(\frac{a+b}{2}\right)}^2\ge ab$
$\Rightarrow \frac{{(a+b)}^2}{4}\ge ab$
$\Rightarrow \frac{ab}{{(a+b)}^2}\le \frac{1}{4}$
Hence,$\frac{ab{c}^2}{{(a+b)}^2}\le \frac{c^2}{4}$