China Girls' Mathematical Olympiad

Solution Using the sine rule, we have that $$\frac{\sin \angle AFD}{\sin \angle FAD}=\frac{AD}{FD}=\frac{AD}{ED}=\frac{\sin \angle AED}{\sin \angle FAD},$$ then $\sin \angle AFD=\sin \angle AED$. So, either $\angle AFD=\angle AED$ or $\angle AFD+\angle AED=180^{\circ }$. If $\angle AFD=\angle AED$, then $△ADF\cong △ADE$, and we get $AF=AE$. Then $△AIF\cong △AIE$, and $\angle AFI=\angle AEI$. So $△AFC\cong △AEB$, then $AC=AB$. It contradicts the condition given. So $\angle AFD+\angle AED=180^{\circ }$, and points $A,F,D$ and $E$ lie on one circle. Then $\angle DEC=\angle DFA>\angle ABC$. Extend $CA$ through $A$ to point $P$ such that $$\angle DPC=\angle B,\text{ then }PC=PE+CE.\stackrel{◯}{1}$$ Since $\angle BFD=\angle PED$ and $FD=ED$, we have that $△BFD\cong △PED$, then $PE=BF=\frac{ac}{a+b}$. Furthermore, $△PCD\sim △BCA$, then $\frac{PC}{BC}=\frac{CD}{CA}$. So $$PC=a\cdot \frac{ba}{b+c}\cdot \frac{1}{b}=\frac{a^2}{b+c}.\stackrel{◯}{2}$$

From ① and ② we get $\frac{a^2}{b+c}=\frac{ac}{a+b}+\frac{ab}{c+a}$, then $\frac{a}{b+c}=\frac{b}{c+a}+\frac{c}{a+b}$.

As to the proof of (2), we have from (1) that $$\begin{array}{rl}a(a+b)(a+c)& =b(b+a)(b+c)+c(c+a)(c+b),\\ a^2(a+b+c)& =b^2(a+b+c)+c^2(a+b+c)+abc\\ & >b^2(a+b+c)+c^2(a+b+c).\end{array}$$ Then $a^2>b^2+c^2$, and that means $\angle BAC>90^{\circ }$.