Question

In $△$ ABC, the bisector of the angle A meets the side BC at D and the circumscribed circle at E,then DE equals

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

Answer 1

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