# Did Galileo use an erroneous geometrical result in 'Two New Sciences'?

In Thm. 4, Prop. 4 of Galileo's 'Two New Sciences' (pg. 187, Crew Translation), Galileo says the following: "From a single point $$B$$ draw the planes $$BA$$ and $$BC$$, having the same length but different inclinations; let $$AE$$ and $$CD$$ be horizontal lines drawn to meet the perpendicular $$BD$$; and let $$BE$$ represent the height of the plane $$AB$$, and $$BD$$ the height of $$BC$$; also let $$BI$$ be a mean proportional to $$BD$$ and $$BE$$; then the ratio of $$BD$$ to $$BI$$ is equal to the square root of the ratio of $$BD$$ to $$BE$$." (See figure)

The claim struck me as odd and I experimented many times with various geometrical figures to see if I could reproduce this result in at least one instance, but I couldn't. What's going on here? • The statement is nothing more than the definition of "mean proportional". Aug 3 '20 at 7:22
• Is line $SI$ perpendicular to $BD$?
– user10963
Aug 3 '20 at 11:24
• @Zenra Yes. Later in the proof Galileo says "Draw IS parallel to DC". The proposition being proved here is "The times of descent along planes of the same length but of different inclinations are to each other in the inverse ratio of the square roots of their heights." Aug 3 '20 at 11:45
• What is the question here? Looks like your "root question" is that of the definition of the "mean proportional" to point I . Aug 3 '20 at 12:07

As $$BI$$ is mean proportional to $$BD$$ and $$BE$$.
$$\begin{array}{l} \Rightarrow \frac{B D}{B I}=\frac{B I}{B E} \\ \Rightarrow \frac{B D}{B I} \times B D=\frac{B I}{B E} \times B D \\ \Rightarrow \frac{B D^{2}}{B I}=\frac{B I \times B D}{B E} \\ \Rightarrow \frac{B D^{2}}{B I^{2}}=\frac{B D}{B E} \end{array}$$
$$Q.E.D$$