## that square root problem

At the BAPL 09 banquet Johnny Ball demonstrated a geometric method to determine a square root. A couple of us couldn't quite believe it at the time. Devon came up with a verbal description next morning.

This is what I make of it.

In the diagram:

BC=1 ⍝ by construction AD=DC ⍝ by construction AD=DE ⍝ by construction Prove: BE=AB*0.5 ⍝ BE is the square root of AB Proof: (BE*2) = (DE*2)-DB*2 ⍝ Pythagorus = (DE+DB)×DE-DB ⍝ difference of 2 squares = AB × 1 .'. BE = AB*0.5 QED

Does anyone know if it's in Euclid?-- PhilLast 2009-06-10 12:05:14