In triangle ABC, the bisector of the angle A meets the side BC at D and the circumscribed circle at E, then DE equal

$\text{ Using the property of angle bisector, we have }\frac{BD}{DC}=\frac{AB}{AC}=\frac{c}{b}$
$\begin{array}{r}\Rightarrow BD+DC=ck+bk=a\\ \Rightarrow k=\frac{a}{b+c}.\end{array}$
$\text{ Also }xy=bc{h}^2\text{ (property of circle) }$
$\begin{array}{r}\Rightarrow x=\left(\frac{2bc\cos \frac{A}{2}}{b+c}\right)\frac{bc{a}^2}{(b+c{)}^2}\\ =\frac{a^2\sec \frac{A}{2}}{2(b+c)}\end{array}$