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Proposition 14

Mon, 2007-01-22 16:26 — ashley

To construct a square equal to t a given rectilineal figure.

Let $ A $ be the given rectilineal figure;

thus it is required to construct a square equal to the rectilineal figure $ A $.

For let there be constructed the rectangluar parallelogram $ BD $ equal to the rectilineal figure $ A $. [Prop. 1.45]

Then, if $ BE $ is equal to $ ED $, that which was enjoined will have been done; for a square $ BD $ has been constructed equal to the rectilineal figure $ A $.

But, if not, one of the straight lines $ BE $, $ ED $ is greater.

Let $ BE $ be greater, and let it be produced to $ F $;

let $ EF $ be made equal to $ ED $, and let $ BF $ be bisected at $ G $.

With centre $ G $ and distance of of the straight lines $ GB $, $ GF $ let the semicircle $ BHF $ be described; let $ DE $ be produced to $ H $, and let $ GH $ be joined.

Then, since the straight line $ BF $ has been cut into equal segments at $ G $, and into unequal segments at $ E $,

the rectangle contained by $ BE $, $ EF $ together with the square on $ EG $ is equal to the square on $ GF $. [Prop. 2.5]

but $ GF $ is equal to $ GH $;

therefore the rectangle $ BE $, $ EF $ together with the square on $ GE $ is equal to the square on $ GH $.

But the squares on $ HE $, $ EG $ are equal to the square on $ GH $; [Prop. 1.47]

therefore the rectangle $ BE $, $ EF $ together with the square on $ GE $ is equal to the squares on $ HE $, $ EG $.

Let the square on $ GE $ be subtracted from each;

therefore the rectangle contained by $ BE $, $ EF $ which remains is equal to the square on $ EH $.

But the rectangle $ BE $, $ EF $ is $ BD $, for $ EF $ is equal to $ ED $;

therefore the parallelogram $ BD $ is equal to the square on $ HE $.

And $ BD $ is equal to the rectilineal figure $ A $.

Therefore the rectilineal figure $ A $ is also equal to the square which can be described on $ EH $.

Therefore a square, namely that which can be described on $ EH $, has been constructed equal to the given rectilineal figure $ A $.

Q.E.F.