equal to BD (4), but the squares of AB and BC are (4) constr. equal to the squares of AC (5), therefore the squares (5) hypoth. of AD and AC are equal, and therefore the lines themselves are equal(6): but also DB and BC are equal, and (6) Cor. 2. the side AB is common to both triangles, therefore the Prop. 46. triangles ABC and ABD are mutually equilateral, and

therefore the angle ABC is equal to the angle ABD(7), (7) Prop. 8. but ABD is a right angle (8), therefore ABC is also a (8) constr. right angle.

F

THE ELEMENTS

OF

EUCLID.

BOOK II.

Plate 2. Fig. 6.

DEFINITIONS.

1. Every rectangle or right angled parallelogram is said to be contained by the two right lines, which contain one of its right angles.

(It is called the rectangle under these two lines.) 2. In any parallelogram either of the parallelograms about the diagonal (EK or OF) with the two complements AG and GD) is called a Gnomon.

PROP. I. THEOR.

If there be two right lines (A and BC), one of which Plate 2. is divided into any number of parts (BD, DE, EC'), Fig. 1. the rectangle under the two lines is equal to the sum of the rectangles under the undivided line (A) and the several parts of the divided line (BD, DE, EC).

From the point B draw BH perpendicular to BC, take on it BF equal to A, and through F draw FL parallel to BC, and draw DG, EK, and CL parallel to BF.

It is evident that the rectangle BL is equal to the rectangles BG, DK and EL; but the rectangle BL is the rectangle under A and BC, for BF is equal to A: and the rectangles BG, DK, and EL are the rectangles under A and BD, A and DE, and A and EC, for each (1) Const. and Prop. of the lines, BF, DG and EK is equal to A (1). 34. B. 1.

Cor. Hence and from prop. 34. B. 1. it appears that the area of a rectangle is found by multiplying its altitude into the base: and from prop. 35. and 36. B.1. it also appears that the area of any parallelogram is found by multiplying its altitude into the base, and from prop. 37. and 38. B. 1. that the area of a triangle is found by multiplying its altitude into half the base.

PROP. II. THEOR.

If a right line (AB) be divided into any two parts Fig. 2. (in C) the square of the whole line is equal to the sum See N. of the rectangles under the whole (AB) and each of the parts (AC, CB).

On AB describe the square ADFB (1), and through (1) Prop.46 C draw CE parallel to AD. The square AF is equal to the rectangles AE and CF. But the rectangle AE is the rectangle under AB and AC, because AD is equal

(2) Constr. to AB (2), and the rectangle CF is the rectangle un (3) Prop.34 der AB and CB, because CE is equal to AB (8).

B. 1.

Fig. 3.

Otherwise thus:

Assume a right line, X, equal to the given line AB. The rectangle under X and AB, or the square of AB (1) hypoth. (1), is equal to the sum of the rectangles under X and (2) Prop.1. AC and under X and CB (2), that is, to the sum of the rectangles under AB and AC, and under AB and CB.

B. 2.

Fig. 4.

PROP. III. THEOR.

If a right line (AB) be divided into any two parts (in C), the rectangle under the whole line (AB) and either part (AC) is equal to the square of that part (AC) together with the rectangle under the parts (AC and CB.)

On AC describe the square ADFC, and through B draw BE parallel to AD until it meet DF produced to E. The rectangle AE is equal to the square ADFC together with the rectangle CE.

But the rectangle AE is the rectangle under AC and (1)constr,& AB, for AD is equal to AC (1), and the square ADFC def.31.B.1. is the square of AC (2), and the rectangle CE is the (2) Constr. rectangle under AC and CB, for CF is equal to AC (1).

Fig. 5.

Otherwise thus:

Assume a right line X equal to AC.

The rectangle under X and AB is equal to the sum of the rectangles under X and AC, and under X and (1) Prop.1. CB (1); but the rectangle under X and AB is the rectangle under AC and AB; and the rectangle under X and AC is the square of AC, and the rectangle under X and CB is the rectangle under AC and CB.

B. 2.

PROP. IV. THEOR.

If a right line (AB) be divided into any two parts Fig. 6. (in O), the square of the whole line is equal to the sum of the squares of the parts and twice the rectangle under the parts.

On AC describe the square ACDB, draw CB, and through O draw OK parallel to AC, cutting CB in G, and through G draw EF parallel to AB.

(1)constr & The square ACDB is equal to the squares EK and Cor. p. 43, OF, together with the rectangle AG and GD.

B 1.

But OF is the square of OB (1), and EK is the (2) constr& square of AO, for EG is equal to AO (2); and AG prop.34.B.1 (3) Def.31. and GD together are equal to double the rectangle B. 1. under the parts, because GD is equal to AG (4), (4) Prop.43 and AG is the rectangle under the parts AO and OB, because OG and OB are equal (3).

Otherwise thus :

B. 1.

Fig. 7.

The square of AB is equal to the sum of the rectangles under AB and AÕ, and under AB and BO (1); but the rectangle under AB and AO is equal to (1) Prop.2. the sum of the rectangle under AO and OB, and the B. 2. square of AO (2), and the rectangle under AB and BO (2) Prop.3. is equal to the sum of the rectangle under AO and OB B. 2. and the square of OB (2), therefore the sum of the rectangles under AB and AO, and under AB and BO, or the square of AB, is equal to the sum of the squares of AO and OB and double the rectangle under AO and OB.

Cor. Hence it is evident that the square of half the line is the fourth part of the square of the whole line, for when the line is bisected the rectangle under the parts is equal to the square of the half line.