Question

In $\Delta ABC,a,b$ and $c$ are the lengths of the sides opposite to the angles $A,B$ and $C$ respectively. The bisector of the $\angle A$ meets the side $BC$ at $D$ and the circumscribed circle at $E$. Then $DE$ equals

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

Answer 1

The correct option is C $\frac{a^2\sec \frac{A}{2}}{2(b+c)}$

Using the property of angle bisector, we have
$\frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}$


$\Rightarrow BD+DC=ck+bk=a\Rightarrow k=\frac{a}{b+c}$
Using intersecting chords theorem,
$xy=bc{k}^2$
$\Rightarrow x=\frac{bc{k}^2}{y}$
$\Rightarrow x=\left(\frac{b+c}{2bc\cos \frac{A}{2}}\right)\frac{bc{a}^2}{(b+c{)}^2}\Rightarrow x=\frac{a^2\sec \frac{A}{2}}{2(b+c)}$