Problem - 3025
In $\triangle{ABC}$, let $AB=c$, $AC=b$, and $\angle{BAC}=\alpha$. If $AD$ bisects $\angle{BAC}$ and intersects $BC$ at $D$, find the length of $AD$.
Because $$S_{\triangle{ABC}} = S_{\triangle{ABD}}+S_{\triangle{ACD}}$$
we have $$\frac{1}{2}\cdot AB \cdot AC \cdot \sin{\angle{BAC}}=\frac{1}{2}\cdot AB \cdot AD \cdot \sin{\frac{1}{2}\angle{BAC}}+\frac{1}{2}\cdot AC \cdot AD \cdot \sin{\frac{1}{2}\angle{BAC}}$$ or $$\frac{1}{2}\cdot b \cdot c\cdot \sin{\alpha} = \frac{1}{2}\cdot b\cdot {AD}\cdot \sin{\frac{\alpha}{2}}+\frac{1}{2}\cdot c\cdot {AD}\cdot \sin{\frac{\alpha}{2}}$$ This leads to $$AD=\frac{bc\sin{\alpha}}{(b+c)\cdot\sin{\frac{\alpha}{2}}}=\boxed{\frac{2bc}{b+c}\cos{\frac{\alpha}{2}}}$$