If \( \frac{a}{4}=\frac{b}{5}=\frac{c}{3} \), prove that: \( \frac{a-b+c}{a+b-c}=\frac{1}{3} \)

Given the equation \( \frac{a}{4} = \frac{b}{5} = \frac{c}{3} \), we can set this equal to a common variable \( k \). Hence, we can express \( a \), \( b \), and \( c \) in terms of \( k \): \[ a = 4k, \quad b = 5k, \quad c = 3k \] Now, we want to prove that: \[ \frac{a - b + c}{a + b - c} = \frac{1}{3} \] First, let's compute \( a - b + c \): \[ a - b + c = 4k - 5k + 3k = (4k - 5k + 3k) = 2k \] Next, we compute \( a + b - c \): \[ a + b - c = 4k + 5k - 3k = (4k + 5k - 3k) = 6k \] Now substituting these results back into the fraction, we have: \[ \frac{a - b + c}{a + b - c} = \frac{2k}{6k} \] Simplifying this expression: \[ = \frac{2}{6} = \frac{1}{3} \] Thus, we have successfully proven that: \[ \frac{a - b + c}{a + b - c} = \frac{1}{3} \] This completes the proof.