PROPOSITION 1. THEOREM.

If there are two straight lines, one of which is divided into any number of parts, the rectangle contained by the two straight lines is equal to the sum of the rectangles contained by the undivided straight line and the several parts of the divided line.

Let P and AB be two straight lines, and let AB be divided into any number of parts AC, CD, DB.

Then shall the rectangle contained by P, AB be equal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB.

Construction.

From A draw AF perp. to AB; and make AG equal to P.

Through G draw GH par1 to AB;

and through C, D, B draw CK, DL, BH par1 to AG.

I. 11.

I. 3.

I. 31.

Proof. Now the fig. AH is made up of the figs. AK, CL, DH, and is therefore equal to their sum;

and of these,

the fig. AH is the rectangle contained by P, AB;

for it is contained by AG, AB; and AG=P:
and the fig. AK is the rectangle contained by P, AC;
for it is contained by AG, AC; and AG=P:
also the fig. CL is the rectangle contained by P, CD;
for it is contained by CK, CD;

and CK = the opp. side AG, and AG=P. I. 34. Similarly the fig. DH is the rectangle contained by P, DB. .. the rectangle contained by P, AB is equal to the sum of the rectangles contained by P, AC, by P, CD, and by P, DB.

Q.E.D..

CORRESPONDING ALGEBRAICAL FORMULA,

In accordance with the principles explained on page 129, the result of this proposition may be written thus:

P. AB P. AC+ P. CD+P.DB.

Now if the line P contains p units of length, and if AC, CD, DB contain a, b, c units respectively,

hence the statement

becomes

then AB=a+b+c;

P. AB P.AC+P. CD+P.DB

p(a+b+c)=pa+pb+pc.

[NOTE. It must be understood that the rule given on page 129, for expressing the area of a rectangle as the product of the lengths of two adjacent sides, implies that those sides are commensurable, that is, that they can be expressed exactly in terms of some common unit.

This however is not always the case. Two straight lines may be so related that it is impossible to divide either of them into equal parts, of which the other contains an exact number. Such lines are said to be incommensurable. Hence if the adjacent sides of a rectangle are incommensurable, we cannot choose any linear unit in terms of which these sides may be exactly expressed; and thus it will be impossible to subdivide the rectangle into squares of unit area, as illustrated in the figure of page 129. We do not here propose to enter further into the subject of incommensurable quantities: it is sufficient to point out that further knowledge of them will convince the student that the area of a rectangle may be expressed to any required degree of accuracy by the product of the lengths of two adjacent sides, whether those lengths are commensurable or not.]

PROPOSITION 2. THEOREM.

If a straight line is divided into any two parts, the square on the whole line is equal to the sum of the rectangles contained by the whole line and each of the parts.

Let the straight line AB be divided at C into the two parts AC, CB.

Then shall the square on AB be equal to the sum of the rectangles contained by AB, AC, and by AB, BC.

Construction. On AB describe the square ADEB.
Through C draw CF par1 to AD.

Proof.

I. 46.

I. 31.

Now the fig. AE is made up of the figs. AF, CE:

and of these,

the fig. AE is the sq. on AB:

Constr. and the fig. AF is the rectangle contained by AB, AC; for it is contained by AD, AC; and AD=AB: also the fig. CE is the rectangle contained by AB, BC; for it is contained by BE, BC; and BE = AB.

the sq. on AB = the sum of the rectangles contained by AB, AC, and by AB, BC.

CORRESPONDING ALGEBRAICAL FORMULA.

The result of this proposition may be written

AB2

AB. AC+ AB. BC.

Q.E.D.

Let AC contain a units of length, and let CB contain b units,

and we have

then AB=a+b units;

(a+b)2=(a+b)a+(a+b)b.

PROPOSITION 3. THEOREM.

If a straight line is divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the square on that part together with the rectangle contained by the two parts.

C

B

Let the straight line AB be divided at C into the two parts AC, CB.

Then shall the rectangle contained by AB, AC be equal to the square on AC together with the rectangle contained by AC, CB. Construction. On AC describe the square AFDC. Through B draw BE par1 to AF, meeting FD produced in E.

I. 46.

I. 31. Proof. Now the fig. AE is made up of the figs. AD, CE; and of these,

the fig. AE is the rectangle contained by AB, AC;

for AF = AC;

Constr.

and the fig. AD is the sq. on AC; also the fig. CE is the rectangle contained by AC, CB;

for CD = AC.

the rectangle contained by AB, AC is equal to the sq. on AC together with the rectangle contained by AC, CB.

CORRESPONDING ALGEBRAICAL FORMULA.

This result may be written AB. AC=AC2+AC . CB.
Let AC, CB contain a and b units of length respectively,

and we have

then AB=a+b units;

(a+b) a = a2+ab.

Q.E.D.

NOTE. It should be observed that Props. 2 and 3 are special cases of Prop. 1.

If a straight line is divided into any two parts, the square on the whole line is equal to the sum of the squares on the two parts together with twice the rectangle contained by the two parts.

Let the straight line AB be divided at C into the two parts AC, CB.

Then shall the sq. on AB be equal to the sum of the sqq. on AC, CB, together with twice the rect. AC, CB.

Construction. On AB describe the square ADEB.

Join BD.

I. 46. Through C draw CF par1 to BE, meeting BD in G. I. 31. Through G draw HGK par1 to AB.

It is first required to shew that the fig. CK is the sq. on CB.

Proof.

Because CF and AD are par1, and BD meets them, .. the ext. angle CGB = the int. opp. angle ADB. I. 29. And since ABAD, being sides of a square;

... CB= CG.

I. 5.

I. 6.

I. 34.

Def. 30.

And the opp. sides of the par CK are equal;
.. the fig. CK is equilateral;
also the angle CBK is a right angle;
CK is a square, and it is described on CB. I. 46, Cor.

Similarly, the fig. HF is the sq. on HG, that is, the

sq. on AC;

for HG the opp. side AC.

I. 34.