63

angle FDB a right angle, for it is equal to the interior and [I. 29. opposite angle ECB ; therefore the remaining angle BFD is half a right angle. Therefore the angle at B is equal to the angle BFD, and the side DF is equal to the side DB. And because AC is equal to CE,

[I. 6. [Construction.

the square on AC is equal to the square on CE; therefore the squares on AC, CE are double of the square on AC.

But the square on AE is equal to the squares on AC, CE, [I. 47. because the angle ACE is a right angle; therefore the square on AE is double of the square on AC. Again, because EG is equal to GF,

[Construction.

the square on EG is equal to the square on GF; therefore the squares on EG, GF are double of the square' on GF.

GF,

[I. 47.

But the square on EF is equal to the squares on EG, because the angle EGF is a right angle ; therefore the square on EF is double of the square on GF. [I. 34. And GF is equal to CD ; therefore the square on EF is double of the square on CD. But it has been shewn that the square on AE is also double of the square on AC.

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

But the square on AF is equal to the squares on AE, [I. 47. EF, because the angle AEF is a right angle. 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, because the angle ADF is a right angle.

[I. 47. Therefore the squares on AD, DE 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, ČD.

Wherefore, if a straight line &c.

Q.E.D.

PROPOSITION 10. THEOREM.

If a straight line be bisected, and produced to any point, the square on the whole line thus produced, and the square on the part of it produced, are together double of the square on half the line bisected and of the square on the line made up of the half and the part produced.

Let the straight line AB be bisected at C, and produced to D: the squares on AD, DB shall be together double of the squares on AC, CD.

From the point Cdraw CE at right angles to AB, [I. 11. and make it equal to AC

or CB;

[I. 3.

and join AE, EB; through
E draw EF parallel to
AB, and through D draw
DF parallel to CE. [I. 31.
Then because the straight

line EFmeets the parallels

B

F

EC, FD, the angles CEF, EFD are together equal to two right angles;

[I. 29.

and therefore the angles BEF, EFD are together less than two right angles.

Therefore the straight lines EB, FD will meet, if produced, towards B, D.

Let them meet at G, and join AG.

[Axiom 12.

Then because AC is equal to CE,

[Construction.

the angle CEA is equal to the angle EAC; and the angle ACE is a right angle;

[I. 5.

[Construction.

therefore each of the angles CEA, EAC is half a right

angle.

[I. 32.

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

Therefore the angle AEB is a right angle.

And because the angle EBC is half a right angle, the angle DBG is also half a right angle, for they are vertically opposite;

[I. 15.

[I. 29.

but the angle BDG is a right angle, because it is equal to the alternate angle DCE; therefore the remaining angle DGB is half a right angle, [1.32.

and is therefore equal to the angle DBG;

therefore also the side BD is equal to the side DG. [I. 6. Again, because the angle EGF is half a right angle, and the angle at F a right angle, for it is equal to the opposite angle ECD;

[I. 34. therefore the remaining angle FEG is half aright angle, [I. 32. and is therefore equal to the angle EGF;

therefore also the side GF is equal to the side FE. [I. 6.

And because EC is equal to CA, the square on EC is

equal to the square on CA ;

therefore the squares on EC, CA are double of the square on CA.

But the square on AE is equal to the squares on EC, CA. [I. 47. Therefore the square on AE is double of the square on AC. Again, because GF is equal to FE, the square on GF is equal to the square on FE;

therefore the squares on GF, FE are double of the square on FE.

But the square on EG is equal to the squares on GF,FE.[1.47. Therefore the square on EG is double of the square on FE. And FE is equal to CD;

[I. 34. therefore the square on EG is double of the square on CD. But it has been shewn that the square on AE is double of the square on AC.

Therefore the squares on AE, EG are double of the squares on AC, CD.

But the square on AG is equal to the squares on AE, EG. [I. 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. 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.

Wherefore, if a straight line &c. Q.E.D.

PROPOSITION 11. PROBLEM.

To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.

Let AB be the given straight line: it is required to divide it into two parts, so that the rectangle contained by the whole and one of the parts may be equal to the square on the other part.

make EF equal to EB;

and on AF describe the AFGH.

AB shall be divided at H so that the rectangle AB, BH is equal to the square on AH.

Produce GH to K.

Then, because the straight line
AC is bisected at E, and pro-

E

duced to F, the rectangle CF, FA, together with the square on AE, is equal to the square on EF. [II. 6. But EF is equal to EB.

[Construction.

Therefore the rectangle CF, FA, together with the square on AE, is equal to the square on EB. But the square on EB is equal to the squares on AE, AB, because the angle EAB is a right angle. [I. 47. Therefore the rectangle CF, FA, together with the square on AE, is equal to the squares on AE, AB.

Take away the square on AE, which is common to both; therefore the remainder, the rectangle CF, FA, is equal to the square on AB. [Axiom 3.

But the figure FK is the rectangle contained by CF, FA, for FG is equal to FA ;

and AD is the square on AB;

therefore FK is equal to AD.

Take away the common part AK, and the remainder FH

is equal to the remainder HD.

[Axiom 3.

But HD is the rectangle contained by AB, BH, for AB is equal to BD;

and FH is the square on AH;

therefore the rectangle AB,BH is equal to the square on AH.

Wherefore the straight line AB is divided at H, so that the rectangle AB, BH is equal to the square on AH. Q.E.F.

PROPOSITION 12. THEOREM.

In obtuse-angled triangles, if a perpendicular be drawn from either of the acute angles to the opposite side produced, 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 the side on which, when produced, the perpendicular falls, and the straight line intercepted without the triangle, between the perpendicular and the obtuse angle.

Let ABC be an obtuse-angled triangle, having the obtuse angle ACB, and from the point A let AD be drawn perpendicular to BC produced: the square on AB shall be greater than the squares on AC, CB, by twice the rectangle BC, CD.

Because the straight line BD is divided into two parts at the point C, the square on BD is equal to the squares on BC, CD, and twice the rectangle BC, CD. [II. 4. To each of these equals add the square on DA.

B

Therefore the squares on BD, DA are equal to the squares on BC, CD, DA, and twice the rectangle BC, CD. [Axiom 2. But the square on BA is equal to the squares on BD, DA, because the angle at D is a right angle; [I. 47. and the square on CA is equal to the squares on CD,DA.[I. 47. Therefore the square on BA is equal to the squares on BC, CA, and twice the rectangle BC, CD;

that is, the square on BA is greater than the squares on BC, CA by twice the rectangle BC, CD.

Wherefore, in obtuse-angled triangles &c.

Q.E.D.