In the construction of this proposition, Euclid has considered only one case or position of the three squares in relation to the sides of the triangle. But there are six different cases or positions of the three squares in relation to the sides of the triangle and to each other, as follows: J. The three squares on the three exterior sides of the triangle. 2. The three squares on the three interior sides of the triangle. 8. The two smaller squares on the exterior sides, and the greater on the interior side. 4. The two smaller squares on the interior sides, and the greater on the exterior side. 5. The greater and one of the smaller squares on the exterior sides, and the other on the interior side. 6. The greater and one of the smaller squares on the interior side, and the other on the exterior side. It will be a useful and instructive exercise to the student to try and adapt the preceding demonstration to all these six different cases or positions of the squares. Exercise 1.-In the figure of the preceding proposition, join GH, FD and KE, and prove that the three triangles, thus formed, are each equal to the triangle ABC, and to one another. Exercise 2.-Prove that the three straight lines A L, FC, and K B, meet in the same point. Exercise 3.-To describe a square equal to two given squares; and, consequently, to any given number of squares. Exercise 4.-To describe a square equal to the difference between any two given squares; and, consequently, to the difference between a given square, and any number of given squares. Exercise 5.-In any triangle, if a perpendicular be drawn from the vertex of any angle to the opposite side, the difference of the squares of the segments of that side is equal to the difference of the squares of the other two sides. DEFINITION.-In any right-angled triangle, the side opposite to the right angle is called the hypothenuse, and the other sides, or legs, are called the base and the perpendicular; that is, if the one leg is the base, the other is the perpendicular, and conversely. If the square described upon one of the sides of a triangle, be equal to the squares described upon the other two sides of it; the angle contained by these two sides is a right angle. Let the square described upon BC, one of the sides of the triangle ABC, be equal to the squares upon the other two sides, AB and A ̊C. Then the angle BAC is a right angle. From the point A draw AD at right angles (I. 11) to AC. Make A D equal (I. 3) to A B, and join DC. Because AD is equal to A B, the square of AD is equal to the square of A B. To each of these equals, add the square of A C. Therefore the squares of AD and A C are equal (Ax. 2) to the squares of AB and AC. But B the squares of AD and AC are equal to the square of DC D (I. 47), because the angle D A C is a right angle. And the square of BC (Hyp.) is equal to the squares of BA and AC. Therefore the square of DC is equal (Ax. 1) to the square of BC, and the side DC to the side BC. Again, because the side AD is equal to the side AB, and AC common to the two triangles DAC, BAC, the two sides DA, AC are equal to the two sides BA, AC, each to each, but the base DC has been proved to be equal to the base BC. Therefore the angle DAC is equal (I. 8) to the angle BAC. But DAC is a right angle. Therefore also BAC is a right angle. Therefore,if the square described upon, &c. Q. E. D. From this proposition is deduced a very useful and practical mode of drawing a perpendicular to a given straight line from any point in the same. With a pair of compasses, mark off from the given point, five equal parts on the given straight line, marking them 1, 2, 3, 4, 5, at their extremities: 0, being marked at the given point. From the point marked 0, with radius equal to 4 of these parts describe an arc of a circle; and from the point marked 3, with radius equal to 5 of these parts, describe an arc of a circle intersecting the former arc. Join the point of intersection of the two arcs with the point marked 0, and the straight line thus drawn will be perpendicular to the given straight line at the given point, as required. For, joining the intersection of the arcs with the point marked 3, a triangle is formed; and the squares of its two sides 3 and 4 are together equal to the square of its side 5; because 9+16=25. Therefore, the sides 3 and 4 contain a right angle, which is opposite to the side 5. The equimultiples of the numbers, 3, 4, and 5, will answer the same purpose equally well. The following propositions may be added to this Book, chiefly as Exercises on the 47th proposition. PROP. A. THEOREM.-In any right angled In the adjacent figure, one method of 2 D G 3 ♡ E E PROP. B. THEOREM.-In any triangle, if squares be described on the base, and on the other two sides; and if the perpendiculars to these sides, drawn from the extremities of the base, be produced to meet the opposite sides of the squares or those sides produced; the two rectangles cut off between these perpendiculars and the sides of the squares drawn from the extremities of the base, are together equal to the square of the base. PROP. C. THEOREM.-If two triangles have two sides of the one equal to two sides of the other, each to each, and the angles opposite to either of the two equal sides be each a right angle, the triangles are equal to one another in all respects. PROP. D. THEOREM.-If two exterior angles of a triangle be bisected, and from the point of intersection of the bisecting lines, a straight line be drawn to the opposite angle, it will bisect that angle. PROP. E. THEOREM.-If from any point within or without any rectilineal figure, perpendiculars be drawn to each side, the sum of the squares of the alternate segments, reckoned from two adjacent angular points, are equal BOOK II. DEFINITIONS. 1. EVERY right-angled parallelogram is called a rectangle, and is said. to be contained by any two of the straight lines which form one of its right angles. A D Thus the rectangle AC is said to be contained by the two sides AB and BC which form the angle ABC; or by any of the other two sides which form the other angles, as BC and CD, CD and D A, or DA and AB; as it may have been described by means of any of these pairs of straight lines. For brevity's sake, the words contained by are generally omitted and a point is placed between the B two sides to indicate that the rectangle is contained by these two sides. A rectangle is said to be contained by two straight lines, when its adjacent sides are equal to those two straight lines, each to each. A square is said to be the square of a given straight line, when A II. B D In every parallelogram, any of the parallelograms about a diagonal, together with the two complements, is called a gnomon. "Thus the parallelogram HG together with the complements AF, F C, is a gnomon, which is more briefly expressed by the letters A G K, or EH C, at the opposite angles of the parallelograms which make up the gnomon." In like manner, the parallelogram EK together with the same two complements, AF, FC, is a gnomon; and is expressed by the letters AKG or CEH, at the opposite angles of he parallelograms which make up the gnomon. H A E D F K B G AXIOMS. The whole is equal to all parts taken together. This axiom, though not laid down by Euclid, is constantly employed in the demonstration of the propositions of this book. II. If two things be equal to one another, both taken together are the double of one of them. This axiom, though not laid down, is assumed by Euclid in Prop. XLII. of Book I. and various other places, but particularly in this book. PROPOSITION 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 rectangle contained by the undivided line, and the several parts of the divided line. Let A and BC be two straight lines; and let BC be divided into several parts at the points D and E. The rectangle contained by the two straight lines A and B C, is equal to the rectangles contained by the straight lines A and B D, A and DE, and A and E C. B D E C From the point B, draw B F at right angles (I. 11) to BC. Make BG equal (I. 3) to A. Through G draw GH parallel (1.31) to BC. And through D, E, and C, draw DK, EL, and CH parallel to BG. Then BH, BK, DL, and E H, are rectangles (I. Def. 36). G F K LH A The rectangle BH is equal II. Ax. 1) to the rectangles BK, DL, and EH. But the rectangle BH is contained by the straight lines A and BC, because GB is equal to A. The rectangle BK is contained by the straight lines A, and BD, because GB is equal to A. The rectangle DL is contained by the straight lines A and DE, because DK is equal to BG (I. 34) and BG is equal to A. In like manner, it is proved that the rectangle EH is contained by the straight lines A and EC. Therefore the rectangle contained by the straight lines A and BC, is equal to the several rectangles contained by the straight lines A and BD, A and DE, and A and EC. Wherefore, if there be two straight lines, &c. Q. E. D. Corollary.-If two straight lines be divided each into any number of parts, the rectangle contained by the two straight lines is equal to the sum of the rectangles contained by the several parts of the one and each of the several parts of the other. PROP. II. THEOREM. If a straight line be divided into any two parts, the rectangles contained by the whole and each of its parts, are together equal to the square of the whole line. Let the straight line AB be divided into any two parts at the point C. The rectangles contained by AB, BC and AB, AC are together equal to the square of AB. Upon AB describe (Î. 46) the square AE. Through C draw CF parallel (I. 31) to AD or BE. A C B D F E The square AE is equal (II. Ax. 1) to the rectangles AF and CE. But AE is the square of AB. The rectangle AF is contained by BA, AC; because DA is equal to AB. The rectangle CE is contained by AB, BC, because BE is equal to AB. Therefore the rectangles contained by AB, AC and AB, BC are equal (Ax. 1) to the square of AB. Wherefore if a straight line, &c. Q. E. D. This proposition is merely a corollary to the first proposition; for, when the two straight lines mentioned in its enunciation are equal, their rectangle is a square. PROP. III. THEOREM. If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts is equal to the rectangle contained by the two parts, together with the square of the foresaid part. Let the straight line A B be divided into any two parts at the point C. The rectangle AB.BC is equal to the rectangle A C.CB, together with the square of BC. Upon BC describe the square CE (I. 46). Produce A C F D B E The rectangle AE is equal (II. Ax. 1) to the rectangles AD and CE. But AE is the rectangle contained by AB, BC, because BE is equal (Const.) to BC. The rectangle AD is contained by AC, CB, because CD is equal to CB. And DB is the square of B C. Therefore the rectangle A B.BC is equal to the rectangle AC.CB, together with the square of B C. If therefore a straight line be divided, &c. Q. E. D. This proposition is only another corollary to the first proposition; for, when the undivided straight line mentioned in its enunciation, is equal to one of the parts of the divided straight line, their rectangle becomes a square. PROP. IV. THEOREM. If a straight line be divided into any two parts, the square of the whole line is equal to the squares of the two parts, together with twice the rectangle contained by the parts. Let the straight line AB be divided into any two parts at C. The square of AB is equal to the squares of AC and CB, together with twice the rectangle AC. CB. Upon A B describe the square AE (I. 46). Join BD. Through C draw CGF parallel to AD or BE (I. 31), and through G draw HGK parallel to AB or DE. But H A C B D F E K Because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (I. 29) to the interior and opposite angle AD B. the angle ADB is equal to the angle ABD (I. 5), because B A is equal to AD, being sides of a square. Therefore the angle CGB is equal (I. Ax. 1) to the angle CBG, and the side BC (I. 6) to the side ĈG. But CB is equal (I. 34) also to GK, and CG to BK. Therefore the figure CK is equilateral. Again, because CG is parallel to BK, and CB meets them, the angles KB C, GCB are equal (I. 29) to two right angles. But the angle KBC is a right angle (Const.) Therefore GCB is a right angle. And the angles CGK, GKB, opposite to these, are also (I. 34) right angles. Therefore CK is rectangular. But it is also equilateral. Therefore it is a square, and it is described upon the side CB. For the same reason HF is a square, and it is described upon the side HG, which is (I. 34) equal to AC. Therefore the figures HF and CK are the squares of AC and CB. Because the complement |