(related to Proposition: 6.19: Ratio of Areas of Similar Triangles)
So it is clear, from this, that if three straight lines are proportional, then as the first is to the third, so the figure (described) on the first (is) to the similar, and similarly described, (figure) on the second. (Which is) the very thing it was required to show.
With the same labels as in Prop. 6.19, if $\gamma=\gamma'$ and $\frac {a}{b}=\frac {b}{b'}$ then
$$\frac{\operatorname{area}\triangle{ABC}}{\operatorname{area}\triangle{DEF}}=\frac {\frac 12ab\sin(\gamma)}{\frac 12bb'\sin(\gamma')}=\frac {a}{b'}=\frac {b^2}{b'^2}.$$
Proofs: 1