\{䟝4. Given $a=\frac{b+c}{\cos B+\cos C}$, determine the shape of $\triangle A B C$.

\begin{array}{l}\text { Solution From the given } a \cos B+a \cos C=(a \cos C \\ +c \cos A)+(a \cos B+b \cos A), \\ (b+c) \cos A=0, \\ A=90^{\circ} . \triangle A B C \text { is a right triangle. }\end{array}