Let \[ AB = \sqrt{BC^2 + AC^2} = \sqrt{m^2 + n^2} \] Therefore, \[ m \sin A + n \sin B = m \cdot \frac{BC}{AB} + n \cdot \frac{AC}{AB} \] \[ = m \cdot \frac{m}{\sqrt{m^2 + n^2}} + n \cdot \frac{n}{\sqrt{m^2 + n^2}} = \frac{m^2}{\sqrt{m^2 + n^2}} + \frac{n^2}{\sqrt{m^2 + n^2}} = \frac{m^2 + n^2}{\sqrt{m^2 + n^2}} = \sqrt{m^2 + n^2} \](proved)