(1) Given $\triangle ABC$ with $AB=c, BC=a, CA=b$. $\triangle ACE\ :\ \cos A=\frac{b}{c+EB}$ and $\triangle ABF\ :\ \cos A=\frac{c}{b+CF}$. Then $EB=\frac{b-c\cos A}{\cos A}$ and $CF=\frac{c-b\cos A}{\cos A}$, adding $EB+CF=\frac{b+c-b\cos A-c\cos A}{\cos A}$. Using in $\triangle ABC\ :\ c=b\cos A+a\cos B$ and $b=a\cos C+c\cos A$, we find $EB+CF=\frac{(b-c\cos A)+(c-b\cos A)}{\cos A}=\frac{a\cos C+a\cos B}{\cos A}\stackrel{given}{=}\frac{a}{\cos A}$. $\triangle AEF\ :\ EF^{2}=(c+EB)^{2}+(b+CF)^{2}-2(c+EB)(b+CF)\cos A$, $EF^{2}=(\frac{b}{\cos A})^{2}+(\frac{c}{\cos A})^{2}-2(\frac{b}{\cos A})(\frac{c}{\cos A})\cos A$, $EF^{2}=\frac{b^{2}+c^{2}-2bc\cos A}{\cos^{2}A}=\frac{a^{2}}{\cos^{2}A}$, then $EF=\frac{a}{\cos A}$.