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Proposition 9

Fri, 2007-01-05 12:17 — ashley

If a straight line be cut into equal and unequal segments, the squares on the unequal segments of the whole are double of the square on the half and of the square on the straight line between the points of section.

For let a straight line $ AB $ be cut into equal segments at $ C $, and into unequal segments at $ D $;

I say that the squares on $ AD $, $ DB $ are double of the squares on $ AC $, $ CD $.

For let $ CE $ be drawn from $ C $ at right angles to $ AB $, and let it be made equal to either $ AC $ or $ CB $;

let $ EA $, $ EB $ be joined,

let $ DF $ be drawn through $ D $ parallel to $ EC $,

and $ FG $ through $ F $ parallel to $ AB $,

and let $ AF $ be joined.

Then, since $ AC $ is equal to $ CE $,

the angle $ EAC $ is also equal to the angle $ AEC $.

And, since the angle at $ C $ is right,

the remaining angles $ EAC $, $ AEC $ are equal to one right angle. [Prop. 1.32]

And they are equal;

therefore each of the angles $ CEA $, $ CAE $ is half a right angle.

For the same reason

each of the angles $ CEB $, $ EBC $ is also half a right angle;

therefore the whole angle $ AEB $ is right.

And, since the angle $ GEF $ is half a right angle.

and the angle $ EGF $ is right, for it is equal to the interior and opposite angle $ ECB $, [Prop. 1.29]

the remaining angle $ EFG $ is half a right angle; [Prop. 1.32]

therefore the angle $ GEF $ is equal to the angle $ EFG $,

so that the side $ EG $ is also equal to $ GF $. [Prop. 1.6]

Again, since the angle at $ B $ is half a right angle,

and the angle $ FDB $ is right, for it is again equal to the interior and opposite angle $ ECB $, [Prop. 1.29]

the remaining angle $ BFD $ is half a right angle; [Prop. 1.32]

therefore the angle at $ B $ is equal to the angle $ DFB $,

so that the side $ FD $ is also equal to the side $ DB $. [Prop. 1.6]

Now, since $ AC $ is equal to $ CE $,

the square on $ AC $ is also equal to the square on $ CE $;

therefore the squares on $ AC $, $ CE $ are double of the square on $ AC $.

But the square on $ EA $ is equal to the squares on $ AC $, $ CE $, for the angle $ ACE $ is right; [Prop. 1.47]

therefore the square on $ EA $ is double of the square on $ AC $.

Again, since $ EG $ is equal to $ GF $,

the square on $ EG $ is also equal to the square on $ GF $;

therefore the squares on $ EG $, $ GF $ are double of the square on $ GF $.

But the square on $ EF $ is equal to the squares on $ EG $, $ GF $;

therefore the square on $ EF $ is double of the square on $ GF $.

But $ GF $ is equal to $ CD $; [Prop. 1.34]

therefore the square on $ EF $ is is double of the square on $ CD $.

But the square on $ EA $ is also double of the square on $ AC $;

therefore the squares on $ AE $, $ EF $ are double of the squares on $ AC $, $ CD $.

And the square on $ AF $ is equal to the squares on $ AE $, $ EF $ for the angle $ AEF $ is right; [Prop. 1.47]

therefore the square on $ AF $ is double of the squares on $ AC $, $ CD $.

But the squares on $ AD $, $ DF $ are equal to the square on $ AF $, for the angle at $ D $ is right; [Prop. 1.47]

therefore the squares on $ AD $, $ DF $ are double of the squares on $ AC $, $ CD $.

And $ DF $ is equal to $ DB $;

therefore the squares on $ AD $, $ DB $ are double of the squares on $ AC $, $ CD $.

Therefore etc.

Q.E.D.