If the angular bisector of the angle A of the triangle ABC meets its circumcircle at E and the opposite side BC at D, then $\mathrm{DE} \cos \frac{\mathrm{A}}{2}=$

• Correct answer
  A

  $\frac{a^2}{2(b+c)}$
• B

  $\frac{b^2}{c+a}$
• C

  $\frac{a}{b+c}$
• D

  $\frac{2 a}{a+b+c}$

Hence, the correct answer is option: 'A'.