Similar triangles are to one another in the duplicate ratio of the corresponding sides.
Proof
Let △ABC∼△DEF with side ratio k=BC/EF.
Construct G on BC so that BG=EF⋅k (i.e.\ a third
proportional, VI.11). By VI.1 the area ratio is BG:EF=k along
one dimension and the side ratio k along the perpendicular, giving
total area ratio k2 in the sense of Definition V.9 (duplicate
ratio).