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 a right 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.

[I. 47.

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

On AB describe the square

ABDC;

[I. 46.

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.

But EF is equal to EB.

[II, 6. [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.

PROPOSITION 13. THEOREM.

+

In every triangle, the square on the side subtending an acute angle, is less than the squares on the sides containing that angle, by twice the rectangle contained by either of these sides, and the straight line intercepted between the perpendicular let fall on it from the opposite angle, and the acute angle.

Let ABC be any triangle, and the angle at B an acute angle; and on BC one of the sides containing it, let fall the perpendicular AD from the opposite angle: the square on AC, opposite to the angle B, shall be less than the squares on CB, BA, by twice the rectangle CB, BD. First, let AD fall within the

triangle ABC.

Then, because the straight line CB is divided into two parts at the point D, the squares on CB, BD are equal to twice the rectangle contained by CB, BD and the square on CD. [II. 7. To each of these equals add the square on DA.

D

Therefore the squares on CB, BD, DA are equal to twice the rectangle CB, BD and the squares on CD, DA. [Ax. 2. But the square on AB is equal to the squares on BD, DA, because the angle BDA is a right angle;

[I. 47. and the square on AC is equal to the squares on CD, DA.[I.47. Therefore the squares on CB, BA are equal to the square on AC and twice the rectangle CB, BD;

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

[II. 12.

and therefore the square on AB is equal to the squares on AC, CB, and twice the rectangle BC, CD. To each of these equals add the square on BC. Therefore the squares on AB, BC are equal to the square on AC, and twice the square on BC, and twice the rectangle BC, CD.

[Axiom 2.

But because BD is divided into two parts at C, the rectangle DB, BC is equal to the rectangle BC, CD and the square on BC;

and the doubles of these are equal,

[II. 3.

that is, twice the rectangle DB, BC is equal to twice the rectangle BC, CD and twice the square on BC.

Therefore the squares on AB, BC are equal to the square on AC, and twice the rectangle DB, BC;

that is, the square on AC alone is less than the squares on AB, BC by twice the rectangle DB, BC.

Lastly, let the side AC be perpendicular to BC.

Then BC is the straight line between the perpendicular and the acute angle at B; and it is manifest, that the squares on AB, BC are equal to the square on AC, and twice the square on BC. [I. 47 and Ax. 2. Wherefore, in every triangle &c. Q.E.D.

4

PROPOSITION 14. PROBLEM.

B

To describe a square that shall be equal to a given rectilineal figure.

Let A be the given rectilineal figure: it is required to describe a square that shall be equal to A.