For let CE be drawn from the point C at right angles to AB [[I. 11](elem.1.11)], and let it be made equal to either AC or CB [[I. 3](elem.1.3)]; let EA , EB be joined; through E let EF be drawn parallel to AD , and through D let FD be drawn parallel to CE . [[I. 31](elem.1.31)]

Then, since a straight line EF falls on the parallel straight lines EC , FD , the angles CEF , EFD are equal to two right angles; [[I. 29](elem.1.29)] therefore the angles FEB , EFD are less than two right angles.

But straight lines produced from angles less than two right angles meet; [[I. Post. 5](elem.1.post.5)] therefore EB , FD , if produced in the direction B , D , will meet.

Let them be produced and meet at G , and let AG be joined.

Then, since AC is equal to CE , the angle EAC is also equal to the angle AEC ; [[I. 5](elem.1.5)] and the angle at C is right; therefore each of the angles EAC , AEC is half a right angle. [[I. 32](elem.1.32)]

For the same reason each of the angles CEB , EBC is also half a right angle; therefore the angle AEB is right.

And, since the angle EBC is half a right angle, the angle DBG is also half a right angle. [[I. 15](elem.1.15)] But the angle BDG is also right, for it is equal to the angle DCE , they being alternate; [[I. 29](elem.1.29)] therefore the remaining angle DGB is half a right angle; [[I. 32](elem.1.32)] therefore the angle DGB is equal to the angle DBG , so that the side BD is also equal to the side GD . [[I. 6](elem.1.6)]

Again, since the angle EGF is half a right angle, and the angle at F is right, for it is equal to the opposite angle, the angle at C , [[I. 34](elem.1.34)] the remaining angle FEG is half a right angle; [[I. 32](elem.1.32)] therefore the angle EGF is equal to the angle FEG , so that the side GF is also equal to the side EF . [[I. 6](elem.1.6)]

Now, since the square on EC is equal to the square on CA , the squares on EC , CA are double of the square on CA .

But the square on EA is equal to the squares on EC , CA ; [[I. 47](elem.1.47)] therefore the square on EA is double of the square on AC . [[C. N. 1](elem.1.c.n.1)]

Again, since FG is equal to EF , the square on FG is also equal to the square on FE ; therefore the squares on GF , FE are double of the square on EF .

But the square on EG is equal to the squares on GF , FE ; [[I. 47](elem.1.47)] therefore the square on EG is double of the square on EF .

And EF is equal to CD ; [[I. 34](elem.1.34)] therefore the square on EG is double of the square on CD . But the square on EA was also proved double of the square on AC ; therefore the squares on AE , EG are double of the squares on AC , CD .

And the square on AG is equal to the squares on AE , EG ; [[I. 47](elem.1.47)] therefore the square on AG is double of the squares on AC , CD . But the squares on AD , DG are equal to the square on AG ; [[I. 47](elem.1.47)] therefore the squares on AD , DG are double of the squares on AC , CD .

And DG is equal to DB ; therefore the squares on AD , DB are double of the squares on AC , CD .

Therefore etc. Q. E. D.

Proposition 11.
To cut a given straight line so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment .

Let AB be the given straight line; thus it is required to cut AB so that the rectangle contained by the whole and one of the segments is equal to the square on the remaining segment.

For let the square ABDC be described on AB ; [[I. 46](elem.1.46)] let AC be bisected at the point E , and let BE be joined; let CA be drawn through to F , and let EF be made equal to BE ; let the square FH be described on AF , and let GH be drawn through to K .

I say that AB has been cut at H so as to make the rectangle contained by AB , BH equal to the square on AH .

For, since the straight line AC has been bisected at E , and FA is added to it, the rectangle contained by CF , FA together with the square on AE is equal to the square on EF . [[II. 6](elem.2.6)]

But EF is equal to EB ; therefore the rectangle CF , FA together with the square on AE is equal to the square on EB .

But the squares on BA , AE are equal to the square on EB , for the angle at A is right; [[I. 47](elem.1.47)] therefore the rectangle CF , FA together with the square on AE is equal to the squares on BA , AE .

Let the square on AE be subtracted from each; therefore the rectangle CF , FA which remains is equal to the square on AB .

Now the rectangle CF , FA is FK , for AF is equal to FG ; and the square on AB is AD ; therefore FK is equal to AD .

Let AK be subtracted from each; therefore FH which remains is equal to HD .

And HD is the rectangle AB , BH , for AB is equal to BD ; and FH is the square on AH ; therefore the rectangle contained by AB , BH is equal to the square on HA . therefore the given straight line AB has been cut at H so as to make the rectangle contained by AB , BH equal to the square on HA . Q. E. F.

Proposition 12.
In obtuse-angled triangles the square on the side subtending the obtuse angle is greater than the squares on the sides containing the obtuse angle by twice the rectangle contained by one of the sides about the obtuse angle , namely that on which the
perpendicular falls , and the straight line cut off outside by the perpendicular towards the obtuse angle .

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](elem.2.4)]

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](elem.1.47)] and the square on AB is equal to the squares on AD , DB ; [[I. 47](elem.1.47)] therefore the square on CB is equal to the squares on CA , AB and twice the rectangle contained by CA , AD ; so that the square on CB is greater than the squares on CA , AB by twice the rectangle contained by CA , AD .

Therefore etc. Q. E. D.

Proposition 13.
In acute-angled triangles the square on the side subtending the acute angle is less than the squares on the sides containing the acute angle by twice the rectangle contained by one of the sides about the acute angle , namely that on which the perpendicular falls , and the straight line cut off within by the perpendicular towards the acutc angle .

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 , the squares on CB , BD are equal to twice the rectangle contained by CB , BD and the square on DC . [[II. 7](elem.2.7)]

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](elem.1.47)] 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.
To construct a square equal to 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 rectangular parallelogram BD equal to the rectilineal figure A . [[I. 45](elem.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 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 , the rectangle contained by BE , EF together with the square on EG is equal to the square on GF . [[II. 5](elem.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 ; [[I. 47](elem.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.

that which was enjoined will have been done, literally would have been done,

γεγονὸς ἂν εἴη τὸ ἐπιταχθέν .

which can be described, expressed by the future passive participle, ἀναγραφησομἑνῳ, ἀναγραφησόμενον .