It's in the attachment
URGENT

Step-by-step explanation:
[tex] \frac{2a}{a + b - c} = \frac{2b}{b + c - a} = \frac{2c}{a + c - b} = k \\ \\ by \: theorem \: on \: equal \: ratios \\ \\ \frac{2a + 2b + 2c}{a + b - c + b + c - a + a + c - b} = k \\ \\ \frac{2(a + b + c)}{a + b + c} = k \\ \\ \therefore \: k = 2[/tex]