Question

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 $DE\cos\frac{A}{2} =$

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

Hint:

This problem elegantly combines three important geometric theorems: the Angle Bisector Theorem, the formula for the length of an angle bisector, and the Power of a Point Theorem (intersecting chords). Recognizing how these apply is key.

Answer 1

The Correct Option is A

Step 1: Use angle bisector properties and Power of a Point.
\[ AD \cdot DE = BD \cdot DC \]
Step 2: Express lengths using triangle sides.
\[ AD = \frac{2bc}{b+c} \cos\frac{A}{2}, BD = \frac{ac}{b+c}, DC = \frac{ab}{b+c} \]
Step 3: Substitute into Power of a Point.
\[ \frac{2bc}{b+c} \cos\frac{A}{2} \cdot DE = \frac{ac}{b+c} \cdot \frac{ab}{b+c} = \frac{a^2 bc}{(b+c)^2} \]
Step 4: Solve for $DE \cos\frac{A{2}$.}
\[ DE \cos\frac{A}{2} = \frac{a^2 bc}{(b+c)^2} \cdot \frac{b+c}{2bc} = \frac{a^2}{2(b+c)} \]