PROPOSITION XXVIII. THEOREM. All the interior angles of a polygon, together with four right angles, are equal to twice as many right angles as the figure has sides. Let ABCDE be any polygon; then all its interior angles A, B, C, D, E, together with four right angles, are equal to twice as many right E angles as the figure has sides. For, from any point, F, within it, draw lines FA, FB, FC, etc., to all the angles. The poly A gon is thus divided into as many triangles as it has sides. Now the sum of the three angles of each of these triangles is equal to two right angles (Pr. 27); therefore the sum of the angles of all the triangles is equal to twice as many right angles as the polygon has sides. But the same angles are equal to the angles of the polygon, together with the angles at the point F, that is, together with four right angles (Pr. 5, Cor. 2). Therefore the angles of the polygon, together with four right angles, are equal to twice as many right angles as the figure has sides. Scholium. When this proposition is applied to concave polygons (Def. 37), each re-entering angle is to be regarded as greater than two right angles. Cor. The sum of the angles of a quadrilateral is four right angles; of a pentagon, six right angles; of a hexagon, eight, etc. PROPOSITION XXIX. THEOREM. If all the sides of any polygon be produced so as to form an exterior angle at each vertex, the sum of these exterior angles will be equal to four right angles. Let all the sides of the polygon ABC, etc., be produced in the same direction; that is, so as to form one exterior angle at each vertex; then will the sum of the exterior angles be equal to four right angles. C For each interior angle ABC, together with its D B adjacent exterior angle ABD, is equal to two right angles (Pr. 2); therefore the sum of all the interior and exterior angles is equal to twice as many right angles as there are sides of the polygon; that is, they are equal to all the interior angles of the polygon, together with four right angles. Hence the sum of the exterior angles must be equal to four right angles (Ax. 3). Therefore, if all the sides, etc. PROPOSITION XXX. THEOREM. The opposite sides and angles of a parallelogram are equal to each other. B Let ABCD be a parallelogram; then will its opposite sides and angles be equal to each other. Draw the diagonal BD; then, because AB is parallel to CD, and BD meets them, the alternate angles ABD, BDC are equal to each other (Pr. 23). Also, because AD is parallel to BC, and BD meets them, the alternate angles BDA, DBC are equal to each other. Hence the two triangles ABD, BDC have two angles, ABD, BDA of the one, equal to two angles, BDC, CBD of the other, each to each, and the side BD included between these equal angles common to the two triangles; therefore their other sides are equal, each to each, and the third angle of the one to the third angle of the other (Pr. 7), viz., the side AB to the side CD, and AD to BC, and the angle BAD equal to the angle BCD. Also, because the angle ABD is equal to the angle BDC, and the angle CBD to the angle BDA, the whole angle ABC is equal to the whole angle ADC. But the angle BAD has been proved equal to the angle BCD; therefore the opposite sides and angles of a parallelogram are equal to each other. Cor. 1. Two parallels, AB, CD, comprehended between two other parallels, AD, BC, are equal; and the diagonal BD divides the parallelogram into two equal triangles. Cor. 2. If one angle of a parallelogram is a right angle, all its angles are right angles, and the figure is a rectangle. A PROPOSITION XXXI. B THEOREM (Converse of Prop. XXX.) If the opposite sides of a quadrilateral are equal, each to each, the equal sides are parallel, and the figure is a parallelogram. Let ABDC be a quadrilateral, having its opposite sides equal to each other, viz., the side AB equal to CD, and AC to BD; then will the equal sides be parallel, and the figure will be a parallelogram. D Draw the diagonal BC; then the triangles ABC, BCD have all the sides of the one equal to the corresponding sides of the other, each to each; therefore the angle ABC is equal to the angle BCD (Pr. 15), and, consequently, the side AB is parallel to CD (Pr. 22). For a like reason, AC is parallel to BD; hence the quadrilateral ABDC is a parallelogram. Therefore, if the opposite sides, etc. If two opposite sides of a quadrilateral are equal and parallel, the other two sides are equal and parallel, and the figure is a parallelogram. Let ABDC be a quadrilateral, having the sides A AB, CD equal and parallel; then will the sides AC, BD be also equal and parallel, and the figure will be a parallelogram. B *D Draw the diagonal BC; then, because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are equal (Pr. 23). Also, because AB is equal to CD, and BC is common to the two triangles ABC, BCD, the two triangles ABC, BCD have two sides and the included angle of the one equal to two sides and the included angle of the other; therefore the side AC is equal to BD (Pr. 6), and the angle ACB to the angle CBD. And, because the straight line BC meets the two straight lines AC, BD, making the alternate angles BCA, CBD equal to each other, AC is parallel to BD (Pr. 22); hence the figure ABDC is a parallelogram. Therefore, if two opposite sides, etc. PROPOSITION XXXIII. THEOREM. The diagonals of every parallelogram bisect each other. Let ABCD be a parallelogram, whose diago- A nals AC, BD intersect each other in E; then will AE be equal to EC, and BE to ED. D E B Because the alternate angles ABE, EDC are equal (Pr. 23), and also the alternate angles EAB, ECD, the triangles ABE, CDE have two angles in the one equal to two angles in the other, each to each, and the included sides AB, CD are also equal; hence the remaining sides are equal, viz., AE to EC, and DE to EB. Therefore the diagonals of every parallelogram, etc. Cor. If the side AB is equal to AD, the triangles AEB, AED have all the sides of the one equal to the corresponding sides of the other, and are consequently equal; hence the angle AEB will equal the angle AED, and therefore the diagonals of a rhombus bisect each other at right angles. BOOK II. RATIO AND PROPORTION. On the Relation of Magnitudes to Numbers. 1. To measure a quantity is to find how many times it contains another quantity of the same kind called the unit. To measure a line is to find how many times it contains another line called the unit of length, or the linear unit. Thus, when a line is said to be fifteen feet in length, it is to be understood that the line has been compared with the unit of length (one foot), and found to contain it fifteen times. The number which expresses how many times a quantity contains the unit is called the numerical measure of that quantity. 2. Ratio is that relation between two quantities which is expressed by the quotient of the first divided by the second. Thus the ratio of 12 to 4 is 12. The ratio of A to B is 4 A The two quantities compared together are called the terms of the ratio; the first is called the antecedent, and the second the consequent. 3. To find the ratio of one quantity to another is to find how many times the first contains the second; i. e., it is to measure the first by the second taken as the unit. If B be taken as the unit of measure, the quotient is the numerical value of A expressed in terms of this unit. B The ratio of two quantities is the same as the ratio of their numerical measures. Thus, if p denotes the unit, and if p is contained m times in A, and n times in B, then A__mp m 4. Two quantities are said to be commensurable when there is a third quantity of the same kind which is contained an exact number of times in each. This third quantity is called the common measure of the two given quantities. Thus the two lines AB, CD are commensurable if there is a third line, MN, which is contained an exact number of times in each; for example, 7 times in AB, and 4 times in CD. The ratio of two, commensurable quantities can therefore be exactly expressed by a number either whole or fractional. The ratio of AB to CD is 7 4 5. Two quantities are said to be incommensurable when they have no common measure. Thus the diagonal and side of a square are said to be incommensurable (see B. IV., Pr. 35); also the circumference and diameter of a circle (see B.VI., Pr. 11). Whether A and B are commensurable or not, their ratio is exA pressed by B 6. To find the numerical ratio of two given straight lines. Sup pose AB and CD are two straight A lines whose numerical ratio is required. E GB L C F D From the greater line, AB, cut off a part equal to the less, CD, as many times as possible; for example, twice with a remainder EB less than CD. From CD cut off a part equal to the remainder EB as often as possible; for example, once with a remainder FD. From the first remainder BE cut off a part equal to FD as often as possible; for example, once with a remainder GB. From the second remainder FD cut off a part equal to the third GB as many times as possible. Continue this process until a remainder is found which is contained an exact number of times in the preceding one. This last remainder will be the common measure of the proposed lines; and, regarding it as the measuring unit, we may easily find the values of the preceding remainders, and at length those of the proposed lines, whence we obtain their ratio in numbers. For example, if we find GB is contained exactly twice in FD, EB=FD+GB=2GB+GB=3GB; AB=2CD+EB=10GB+3GB=13GB. The ratio of the two lines AB, CD is therefore equal to that of 7. It is possible that, however far this operation is continued, we may never find a remainder which is contained an exact number of times in the preceding one. In such a case, the two quan |