$$\frac{a}{b}=\frac{c-e}{d-f}\ \text{from}\ \frac{a}{b}=\frac{c}{d}=\frac{e}{f}$$
I thought this was componendo-dividendo, but it's not. This is something I've never seen before. I tried to prove it by subtracting $1$ from both sides, didn't work. No path seems to work. I found this in my book, but they didn't provide any justification. They just used it. How can I prove this?
Given that $$ {a \over b} = {c \over d} = {e \over f} = k \ \ \ (\mbox{say}) \tag{1} $$
Then we have $$ a = b k, \ \ c = d k, \ \ e = f k \tag{2} $$
Note that $$ c - e = (d - f) k $$
Thus, $$ {c - e \over d - f} = k \tag{3} $$
From (1) and (3), we conclude that $$ {a \over b} = {c - e \over d - f} $$