Let the number AB be parts of the number C, and another number DE be the same parts of another number F.
I say that, alternately, C is the same parts or part of F that AB is of DE.
Since DE is the same parts of F as AB is of C, therefore F is the same parts of DE as C is of AB.
Divide AB into the parts of C, namely AG and GB, and divide DE into the parts of F, namely DH and HE. Then the multitude of AG and GB equals the multitude of DH and HE.
Now since DH is the same part of F as AG is of C, therefore, alternately, C is the same part or the same parts of F as AG is of DH.
For the same reason, C is the same part or the same parts of F as GB is of HE, so that, in addition, C is the same part or the same parts of F as AB is of DE.
Therefore, if a number is parts of a number, and another is the same parts of another, then alternately, whatever part of parts the first is of the third, the same part, or the same parts, the second is of the fourth.
The sample value taken for m/n in the proof is 2/3.