Question

In a $\Delta ABC$, the bisector of the angle A meets the side BC in D and the circumcircle in E, then DE is equal to
(following usual notations)

• A. $\frac{a^{2}se{c}^2\frac{A}{2}}{2(b+c)}$
• B. $\frac{a^{2}sec\frac{A}{2}}{2(b+c)}$
• C. $\frac{a^{2}sec\frac{A}{2}}{4(b+c{)}^2}$
• D. none of these

Answer 1

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

By property of intersecting chords in a circle,
$AD\times DE=BD\times DC$
By Angle bisector property,
$AD=\frac{2bc}{b+c}cos\left(\frac{A}{2}\right)$
$BD=\frac{ac}{b+c}$
$DC=\frac{ab}{b+c}$
$\therefore DE=\frac{(b+c)sec\frac{A}{2}}{2}\times \frac{a^2}{(b+c{)}^2}=\frac{a^{2}sec\frac{A}{2}}{2(b+c)}$