From this proposition is deduced a very useful and practical mode of draxing 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 che straight Ime 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 8 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 triangle, the squares A F and AK, described on the two sides which contain the right angle BAC, may be so divided by straight lines, that the parts may be applied to the square BE described upon the side opposite the right angle, and made exactly to cover it. In the adjacent figure, one method of proving this ingenious and important theorem by ocular demonstration is suggested; various other methods may be proposed, but that which requires the fewest number of parts, is most probably the best. The theoretical demonstration will form a highly useful exercise to the student. G H 5 2 D 5 K 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 11. DEFINITIONS. I. 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. D Thus the rectangle A C is said to be contained by the two sides A B and BC which form the angle ABC; or by any of the other two sides A 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 it is described upon that straight line; or, a straight line equal to it. Thus the square AC is said to be the square of AB, BC, CD, or DA; as it may have been described on either of these straight lines; it may also be said to be the square of any straight line equal to one of these straight lines. 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 A F, F C, is a gnomon, which is more briefly expressed by the letters A G K, or E H 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, A F, FC, is a gnomon; and is expressed by the letters AKG or CEH, at the opposite angles of the parallelograms which make up the gnomon. AXIOMS. The whole is equal to all parts taken together. H A E BG D K This axiom, though not laid down by Euclid, is constantly employed in the de monstration 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 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; 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 D E, and A and E C. B D EC K LH 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 (I. 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). 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. F A 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. 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 iine. 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 (I. 46) the square AE. Through C draw CF parallel (I. 31) to AD or BE. A C B Ꭰ F К The square AE is equal (II. Ax. 1) to the rectangles AF and CE. But A E 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 A B. 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. 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 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, F because CD is equal to CB. And DB is the square of B C. Therefore the rectangle A B.BC is equal to the rectangie AC.C B, 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 part 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 A B 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. H A Because CF is parallel to AD, and BD falls upon them, the exterior angle BGC is equal (I. 29) D to the interior and opposite angle AD B. But C B E K the angle A D B is equal to the angle ABD (I. 5), because BA 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 AG is equal (I. 43) to the complement GE; and AG is the rectangle contained by AC. CB, for GC is equal to CB. Therefore GE is also equal to the rectangle AC.CB. Wherefore AG and GE are equal to twice the rectangle AC.CB. Also HF and CK are the squares of AC and CB. Therefore the four figures HF, CK, AG and GE, are equal to the squares of AC and CB, with twice the rectangle AC. CB. But the figures HF, CK, AG and GE make up the whole figure AE, which is the square of AB. Therefore the square of AB is equal to the squares of AC and CB, and twice the rectangle AC. CB. Wherefore, if a straight line be divided, &c. Q. E. D. COR. 1.-From the demonstration, it is manifest, that the parallelograms about the diameter of a square, are likewise squares. Corollary 2.-The square of any straight line is equal to four times the square of half the straight line. The demonstration of the preceding proposition might have been shortened by the application of some corollaries to the propositions in Book I. It will be a useful exercise for the student to discover this abridgment himself. The following demonstration is founded on the preceding propositions of this book. Because AB is divided into any two parts at C, the square of AB is equal (II. 2) to the two rectangles AB.AC, and AB.BC. But because AB is divided into any two parts at C, the rectangle AB. AC is equal (II. 8) to the rectangle A C.CB, and the square of AC; also the rectangle AB.BC is equal to the rectangle AC.CB, and the square of CB. Adding these equals, the rectangles AB.AC and AB.BC, are together equal to twice the rectangle AC.CB, and the squares of AC and CB. But it has been proved that the square of AB is equal to the same two rectangles. Therefore the square of AB is equal I. Ax D to the squares of AC and CB, and twice the rectangle AC.CB If a straight line be divided into two equal parts, and also into two unequa. parts; the rectangle contained by the unequal parts, together with the square of the line between the points of section (or division), w equal to the square of half the line. Let the straight line AB be divided into two equal parts at the point C, and into two unequal parts at the point D. The rectangle AD,DB, together with the square of CD, is equal to the square of CB. Upon CB describe (I. 46) the square CF. Join BE. Through D draw DHG parallel to CE or BF (L. 31); and through H draw KLM parallel to CB or EF. Also through A draw AK parallel to CL or BM. K A C D B H M E G F Because the complement CH is equal (I. 43) to the complement HF. To each of them, add DM. Therefore the whole CM is equal to the whole DF. But CM is equal to AL (I. 36) because AC is equal to CB. Therefore AL is equal to DF. To each of these equals, add CH. Therefore the whole AH is equal to DF and CH; that is, to the gnomon CMG. But AH is the rectangle AD.DB, for DH is equal to DB. Therefore the gnomon CMG is equal to the rectangle AD.DB. To each of these equals add LG, which is equal (II. 4. Cor.) to the square of CD. Therefore the gnomon CMG, together with LG. |