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 $\triangle ADF \cong \triangle ADE$, and we get $AF = AE$. Then $\triangle AIF \cong \triangle AIE$, and $\angle AFI = \angle AEI$. So $\triangle AFC \cong \triangle AEB$, 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
$$\angle DPC = \angle B, \text{ then } PC = PE + CE. \quad \textcircled{1}$$
Since $\angle BFD = \angle PED$ and $FD = ED$, we have that $\triangle BFD \cong \triangle PED$, then $PE = BF = \frac{ac}{a+b}$. Furthermore, $\triangle PCD \sim \triangle BCA$, then $\frac{PC}{BC} = \frac{CD}{CA}$. So
$$PC = a \cdot \frac{ba}{b+c} \cdot \frac{1}{b} = \frac{a^2}{b+c}. \quad \textcircled{2}$$

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
$$\begin{aligned} a(a+b)(a+c) &= b(b+a)(b+c) + c(c+a)(c+b), \\ a^2(a+b+c) &= b^2(a+b+c) + c^2(a+b+c) + abc \\ &> b^2(a+b+c) + c^2(a+b+c). \end{aligned}$$
Then $a^2 > b^2 + c^2$, and that means $\angle BAC > 90^\circ$.