Let \(bcx = cay = abz = k\), where \(k \ne 0\) \(\therefore x = \frac{k}{bc},\ y = \frac{k}{ca},\ z = \frac{k}{ab}\) First expression: \[ \frac{ax + by}{a^2 + b^2} = \frac{a \cdot \frac{k}{bc} + b \cdot \frac{k}{ca}}{a^2 + b^2} = \frac{\frac{ak}{bc} + \frac{bk}{ca}}{a^2 + b^2} = \frac{\frac{a^2k + b^2k}{abc}}{a^2 + b^2} = \frac{k(a^2 + b^2)}{abc(a^2 + b^2)} = \frac{k}{abc} \] Second expression: \[ \frac{by + cz}{b^2 + c^2} = \frac{b \cdot \frac{k}{ca} + c \cdot \frac{k}{ab}}{b^2 + c^2} = \frac{\frac{bk}{ca} + \frac{ck}{ab}}{b^2 + c^2} = \frac{\frac{b^2k + c^2k}{abc}}{b^2 + c^2} = \frac{k(b^2 + c^2)}{abc(b^2 + c^2)} = \frac{k}{abc} \] Third expression \[ \frac{cz + ax}{c^2 + a^2} = \frac{c \cdot \frac{k}{ab} + a \cdot \frac{k}{bc}}{c^2 + a^2} = \frac{\frac{ck}{ab} + \frac{ak}{bc}}{c^2 + a^2} = \frac{\frac{c^2k + a^2k}{abc}}{c^2 + a^2} = \frac{k(c^2 + a^2)}{abc(c^2 + a^2)} = \frac{k}{abc} \] Therefore, \[ \frac{ax + by}{a^2 + b^2} = \frac{by + cz}{b^2 + c^2} = \frac{cz + ax}{c^2 + a^2} \quad \text{(Proved)} \]