Problem
GEO-B3-M06-P004 Length of an Angle Bisector
#4
★☆☆☆☆ Level 1 of 5
In triangle \(ABC\), angle bisector \(AD\) meets \(BC\). Prove that \(AD=\frac{2AB\cdot AC\cos\frac A2}{AB+AC}\).
Inspired by Prasolov trigonometric geometry method
C. Hint 1. Express area \(ABC\) as the sum of areas \(ABD\) and \(ACD\).
D. Hint 2. Use the area formula with \(AD\) and angles \(\frac A2\).
The area of the whole triangle is \(\frac12 AB\cdot AC\sin A=AB\cdot AC\sin\frac A2\cos\frac A2\).
On the other hand, \(S_{ABC}=S_{ABD}+S_{ACD}=\frac12 AD\cdot AB\sin\frac A2+\frac12 AD\cdot AC\sin\frac A2=\frac12 AD(AB+AC)\sin\frac A2\).
Canceling \(\sin\frac A2\), we get \(AD=\frac{2AB\cdot AC\cos\frac A2}{AB+AC}\).
The problem shows that not every length is best found by the cosine rule.