Both groups were equally amazed when told that it would make no difference.

The is the statement that the sum of (the areas of) the two small squares equals (the area of) the big one.

The proof below is a somewhat shortened version of the original Euclidean proof as it appears in Sir Thomas Heath's translation. This is because, and ∠BAF = ∠BAC ∠CAF = ∠CAB ∠BAE = ∠CAE.

ΔABF has base AF and the altitude from B equal to AC.

I also have dynamic sum measures, and percentage of total spend per supplier, all dynamic with year selected.

The problem is when I try to accumulate the percentages to create a pareto diagram as I did above (I followed this receipe to create the "static" pareto https://powerbi.tips/2016/10/pareto-charting/).

Thus the area of ΔAEC equals half that of the rectangle AELM.

Then he asked, "Suppose these three squares were made of beaten gold, and you were offered either the one large square or the two small squares. " Interestingly enough, about half the class opted for the one large square and half for the two small squares.

In all likelihood, Loomis drew inspiration from a series of short articles in published by B. (In all, there were 100 "shorthand" proofs.) I must admit that, concerning the existence of a trigonometric proof, I have been siding with with Elisha Loomis until very recently, i.e., until I was informed of Proof #84.

Actually, for some people it came as a surprise that anybody could doubt the existence of trigonometric proofs, so more of them have eventaully found their way to these pages.

Rather inadvertently, it pops up in several Sangaku problems. The proof has been illustrated by an award winning Java applet written by Jim Morey.

I include it on a separate page with Jim's kind permission.

I.e a A supplier is one of the supplier in the interval of 0-80% accumulated sales (ABC-analysis).

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