== 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.

{{attachment:pythag.jpg}}

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 <<DateTime(2009-06-10T12:05:14Z)>>
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CategoryBapla09 CategoryMaths