In the book "Fundamentals groups and covering spaces" by Elon Lages Lima:
I can't prove equality. I only get the following (and similar results) $|f(x)-g(x)|=|f(x)-tf(x)+tf(x)-g(x)+tg(x)-tg(x)|=|(1-t)f(x)+tg(x)+tf(x)-g(x)(1+t)|=|0+tf(x)-g(x)(1+t)|=|tf(x)-g(x)(1+t)|$ because $0=(1-t)(x)+tg(x)$ some $t\in [0,1]$. How can I demonstrate that equality? I see it intuitively but not algebraically. Why $|f(x)-g(x)|=|f(x)|+|g(x)|$?
The point 0 is in the segment $[f(x),g(x)]$ means you have $f(x),0,g(x)$ in that order, so the distance $d(f(x),g(x))$ is the sum of the two distances $d(f(x),0)$ and $(d(0,g(x))$ giving you the result.