Label the top left hand corner of the paper A. Label the bottom left hand corner B and the bottom right hand corner C.
There are three fold lines in the paper after folding twice as instructed. Label the left hand end of the top fold D (directly under A), and left hand end of the bottom fold F (directly over B). Label the other end of the bottom fold G (at the edge of the paper). Join AG to make right triangle AFG. Label the other end of the top fold E where it meets hypotenuse AG. (If you provide a labelled drawing based on the above description, you won’t need the description!)
The triangles ADE and AFG are similar, so AD/AF=DE/FG.
Let y=AB and x=FG (the dimensions of the paper). AD=y/4 because the paper was folded twice, and AF=¾AB=3y/4. Therefore, AD/AF=⅓=DE/x, making DE=x/3, and x=3DE. But FG=BC=x, so the vertical folds divide the paper into thirds.
We have used the proportionalities (constant ratios) of the sides of the similar triangles to prove the proposition.
(The specially constructed diagonal AG transposes the vertical spacing of the folds into horizontal divisions—this is the principle behind the method.)