Join BF. The square of BF, or of its equal BA, is equal to the squares of BE and EF (47. 1), therefore the square of EF is equal to the excess of the square of AB above that of BE or C.

PROP. XLVIII. THEOR.

If the square of one side (BC), of a triangle (ABC), be equal to the squares of the other two sides (AB, AC); the angle (BAC) contained by these two sides, is a right angle.

From the point A, draw AD, perpendicular to one of the sides AC containing the right angle, and equal to the other AB, and join DC.

Because DA is equal to AB, the square of DA is equal to the square of AB (Cor. 3. 34. 1), adding to each the square of AC, the squares of AD and AC are equal to the squares of BA and AC; but the square of DC is equal to the squares of DA and AC (47. 1), and the square of BC is equal to the squares of BA and AC (Hyp.), therefore the squares of DC and BC are equal, and therefore the right line DC is equal to BC (Cor. 1. 46. 1); whence, the right lines DA, AC being severally equal to BA, AC, the angles DAC, BAC are equal (8. 1); but the angle DAC is a right angle (Constr.), therefore the angle BAC is a right angle.

BOOK II.

DEFINITIONS.

1. A Rectangle, is a right angled parallelogram, (as ABCD,) and is said to be contained under any two of the right lines (AB, AD), which make one of its right angles.

The rectangle under two right lines, is that contained by these right lines, or by right lines equal to them, which is equal by Cor. 3. 34. 1.

When a rectangle is denoted by three letters, the middle one is an extreme of both the right lines containing it, the other two letters being at the other extremes of these right lines; thus, the rectangle DAB, denotes the rectangle contained by the right lines DA, AB.

2. In any parallelogram, either of the parallelograms (HG or EK), which are about a diagonal, together with the two complements (AF and FC), is called a Gnomon.

A gnomon is generally denoted by three letters, which are at the opposite angles of the parallelograms which make the gnomon; thus, the gnomon, made by the parallelogram EK with the complements, is denoted by the leters AKG or HEC.

PROPOSITION I. THEOREM.

If there be two right lines (A and BC), whereof one (BC) 5 divided into any number of parts (BD, DE, EČ); the rectangle under these right lines, is equal to the rectangles under the undivided line (A), and all the parts of the divided one (BD, DE and EC).

From B, draw BF perpendicular to BC (11. 1), on which take BG equal to A (3. 1), through G, draw GH parallel to BC, and through D, E, C, draw DK, EL, CH parallel to BG (31. 1), meeting GH in K, L, H.

The rectangle BH is equal to all its parts the rectangles BK, DL, EH (Cor. Ax. 8), but, because BG is equal to A (Constr.), the rectangle BH is equal to the rectangle under BC and A (Cor. 3. 34. 1), and because BG is equal to A (Constr.), and DK, EL each equal to BG (34. 1), and therefore to A (Constr. and Ax. 1), the parallelograms BK, DL, EH are equal to those under BD and A, DE and A, and EC and A.

PROP. II. THEOR.

If a right line (AB), be divided into any two parts (AC, CB); the square of the whole (AB), is equal to the rectangles (BAC, ABC), under the whole (AB), and each of the parts (AC, CB)

On AB describe the square AE (46. 1), and through C, draw CF parallel to AD [31. 1].

The square AE of AB, is equal to the rectangles AF, CE [Cor. Ax. 8]; but the rectangle AF is equal to the rectangle under AB, AC, because AD is equal to AB; and the rectangle CE is equal to

D

F

E

the rectangle under AB, CB, because BE is equal to AB.

Otherwise.

Take the right line Z equal to AB [3. 1]. The rectangle under Z and AB, or, which is equal [Cor. 3. 34. 1], the square of AB, is equal to the rectangles under Z and AC, and under Z and CB [1. 2]; or, which is equal [Cor. 3. 34. 1], to the rectangles under AB, AC, and under AB, CB.

PROP. III. THEOR.

If a right line (AC), be divided into any two parts (AB, BC); the rectangle (CAB), under the whole (AC), and one of the parts (AB), is equal to the rectangle (ABC) under the parts, with the square of the first mentioned part (AB).

On AB describe the square AE [46. 1], and through C draw CF parallel to BE, meeting DE produced in F.

The rectangle AF, is equal to the square AE, with the rectangle BF; but AF is the rectangle under AC, AB, because AD is equal to AB [Constr. and Def. 36. 1], AE is the square of AB (Constr.), and BF is the rectangle under AB, BC, because BE is equal to AB.

Otherwise.

Take Z equal to AB [3. 1]. The rectangle under Z and AC, is equal to the rectangles under Z and AB, and under Z and BC [1.2]; but, because Z is equal to AB [Constr.], the rectangle under Z and AC, is the rectangle CAB, the rectangle under Z and AB, the square of AB, and the rectangle under Z and BC, the rectangle ABC.

Scholium.-This proposition and the preceding, may be virtually comprised in one, namely,

The square of the greater, of two unequal right lines, is greater than the rectangle under the greater and their difference, and the square of either of two right lines, is less than the rectangle under it and their sum, by the rectangle under the same right lines; as is manifest, by supposing, in both these propositions, the right lines to be AB, BC.

PROP. IV. THEOR.

If a right line (AB), be cut into any two parts (AC, CB), the square of the whole (AB), is equal to the squares of the parts (AC, CB), with double the rectangle (ACB) under the parts.

On AB describe the square ABED [46. 1], draw BD, through C, draw CF parallel to AD, meeting BD in G, and through G, draw HK parallel to AB.

The square AE, is equal to CK, HF, with the rectangles AG, GE.

But CK is the square of CB [Constr. and Cor. 2. 46. 1]; and, because HG is equal to AC [34. 1], HF is the square of AC [Constr. and Cor. 2. 46. 1]; and AG is the rectangle under the parts AC, CB, because CG is equal to CB [Def. 36. 1]; whence, the complements AG, GE being equal [43. 1], GE is also the rectangle under AC, CB.

Otherwise.

The square of AB is equal to the rectangles BAC and ABC [2.2]; but the rectangle BAC is equal to the rectangle ACB with the square of AC [3. 2], and the rectangle ABC is equal to the rectangle ACB, with the square of CB [3. 2], therefore the square of AB, which is equal to the rectangles BAC and