Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

Let ABC be an obtuse-angled triangle having the angle BAC obtuse, and let BD be drawn from the point B perpendicular to CA produced;

I say that the square on BC is greater than the squares on BA, AC by twice the rectangle contained by CA, AD.

For, since the straight line CD has been cut at random at the point A, the square on DC is equal to the squares on CA, AD and twice the rectangle contained by CA, AD. [II. 4]

Let the square on DB be added to each; therefore the squares on CD, DB are equal to the squares on CA, AD, DB and twice the rectangle CA, AD.

But the square on CB is equal to the squares on CD, DB, for the angle at D is right; [I. 47]

Therefore etc. Q. E. D.

Proposition 13.

Let ABC be an acute-angled triangle having the angle at B acute, and let AD be drawn from the point A perpendicular to BC;

I say that the square on AC is less than the squares on CB, BA by twice the rectangle contained by CB, BD.

For, since the straight line CB has been cut at random at D,

Let the square on DA be added to each; therefore the squares on CB, BD, DA are equal to twice the rectangle contained by CB, BD and the squares on AD, DC.

But the square on AB is equal to the squares on BD, DA, for the angle at D is right; [I. 47] and the square on AC is equal to the squares on AD, DC; therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD,

so that the square on AC alone is less than the squares on CB, BA by twice the rectangle contained by CB, BD.

Therefore etc. Q. E. D.

Proposition 14.

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 rectangular parallelogram BD equal to the rectilineal figure A. [I. 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 one 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,

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; [I. 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;