BOOK II.

DEFINITIONS.

I.

A Rectangle, or right-angled parallelogram, is said to be contained by any two of the straight lines which contain one of the right angles.

1. The rectangle ABCD is said to be contained by AB and BC, or by BC and CD, or by CD and DA, or by DA and AB.

2. The expression 'the rectangle AB, BC,' is allowed to be used instead of the larger one 'the rectangle contained by AB and BC.

3. The rectangle is often referred to by the two letters standing at its opposite angles, as, in the above, rect. AC, or rect. DB, or by AC or BD only.

II.

In every parallelogram the figure composed of either of the parallelograms about the diameter, together with the two complements, is called a gnomon.

1. In the parallelogram ABCD, the parallelogram EK about the diameter, together with the complements AF and FC, forms the gnomon AKG.

2. Similarly the parallelogram HG, about the diameter, together with the complements AF and FC, forms the gnomon EHC.

3. The gnomon is briefly expressed by the letters at the opposite angles of the parallelograms composing it. Thus the first gnomon AKG, composed of EK, AF, and FC, may be termed the gnomon AKG or HEC. Also the second gnomon EHC, composed of HG, AF, and FC, may be termed the gnomon EHC or AGK.

PROP. I. THEOREM.

If there be 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 rectangles contained by the undivided line and the several parts of the divided line.

Let A and BC be two straight lines, one of which, BC, is divided into parts in D and E.

CONSTRUCTION.-1. From B draw BF at right angles to

BC (I. 11), and cut off BGA (I. 3).

2. Through G draw GH parallel to BC, meeting CH drawn parallel to BG (I. 31).

3. Through D and E draw DK and EL parallel to BG or CH (I. 31).

PROOF.-1. It is evident that the figure BH = the sum of the figures BK, DL, and EH.

2. But because BH is contained by BG and BC, of which BGA (cons.), therefore BH is the rectangle contained by A and BC, the two given straight lines.

3. Similarly BK, DL, and EH are respectively the rectangles contained by BG and BD, by DK and DE, and by EL and EC-i.e. the rectangles contained by A and BD, by A and DE, and by A and EC (cons. and I. 34).

Therefore, it is proved, as required, that

The rectangle contained by the two straight lines A and BC: the rectangles contained by A and BD. by A and DE, and by A and EC.

Wherefore,

If there be two straight lines, &c.

Q. E. D.

Exercise.

Prove the above Proposition with MN and O, the given straight lines, MN being divided into parts in P, Q, and R.

PROP. II. THEOREM.

If a straight line be divided into any two parts, the rectangles contained by the whole line and each of its parts are together equal to the square on the whole line.

C.

Let AB be a straight line divided into any two parts in

CONSTRUCTION.-1. On AB describe the square ADEB

(I. 46).

2. Through C draw CF parallel to AD or BE (I. 31).

PROOF.-1. It is evident that the figure AE = the sum of the figures AF and CE.

2. But because AE is the square on AB (cons.), and because AF and CE are respectively the rectangles AD, AC, and CF, CB, i.e. the rectangles AB, AC, and AB, CB (cons., and I. 34),