parallel; therefore the angle AGH is not unequal to the angle GHD, that is, it is equal to it. But the angle AGH is equal (I. 15.) to the angle EGB; therefore, likewise EGB is equal (I. ax. 1.) to GHD. Add to each of these the angle BGH; therefore the angles EGB, BGH are equal (I. ax. 2.) to the angles BGH, GHD; but EGB, BGH are equal (I. 13.) to two right angles ; therefore also BGH, GHD are equal to two right angles. Wherefore, if a straight line, &c. Cor. ). Hence, if there be two parallels, a straight line which intersects one of them, will intersect the other, if continually produced. Thus, if any straight line be drawn through G, intersecting AB, the angles which it and CD make on one side with GA will be together less than two right angles, and therefore (I. ax, 12.) the lines will meet. Cor. 2. Hence also, two straight lines which intersect each other are not both parallel to the same straight line. PROP. XXX. THEOR. Two straight lines which are not in the same straight line, and which are parallel to another straight line, are parallel to one another. Let the straight lines AB, CD be each of them parallel to the straight line EF; AB is also parallel to CD. Let the straight line LH cut AB, CD, EF; and because LH cuts the parallel straight lines AB, EF, the angle LGB is equal (1.29. part 2.) to the angle LHF. Again, because the straight line LH cuts the parallel straight lines CD, EF, the angle LKD is equal (I. 29. part 2.) to the angle LÅF; and it was shown that the angle LGB is equal to LHF; therefore also LGB is equal (I. ax. I.) to LKD, the interior and remote angle on the same side of LH: therefore AB is parallel (1.28. part 1.) to CD. Wherefore two straight lines, &c. A B K D F PROP. XXXI. PROB. To draw a straight line parallel to a given straight line through a given point without it. Let AB be the given straight line, and C the given point; it is required to draw a straight line through C, parallel to AB. In AB take any point D, and join CD; at the point C, in the straight line CD, make (I. 23.) the angle DCE equal to CDB; and produce the straight line EC to any point F. E F А D Because (const.) the straight line CD, which meets the two straight lines AB, EF, makes the alternate angles ECD, CDB equal to one another, EF is parallel (I. 27.) to AB. Therefore the straight line ECF is drawn through the given point C parallel to the given straight line AB: which was to be done. A E B с D PROP. XXXII. THEOR. If a side of any triangle be produced, (1.) the exterior angle is equal to the two interior and remote angles ; and (2.) tħe three interior angles of every triangle are together equal to two right angles. Let ABC be a triangle, and let one of its sides BC be produced to D: (1.) the exterior angle ACD is equal to the two interior and remote angles CAB, ABC; and (2.) the three interior angles, ABC, BCA, CAB, are together equal to two right angles. Through the point C draw (I. 31.) CE parallel to the straight line AB. Then, because AB is parallel to EC, and AC falls upon them, the alternate angles BAC, ACE are (I. 29. part I.) equal. Again, because AB is parallel to EC, and BD falls upon them, the exterior angle ECD is equal (I. 29. part 2.) to the interior and remote angle ABC: but the angle ACE was shown to be equal to the angle BAC; therefore the whole exterior angle ACD is equal (I. ax. 2.) to the two interior and remote angles CAB, ABC. To these equals add the angle ACB, and the angles ACD, ACB are equal (I. ax. 2.) to the three angles CBA, BÁC, ACB; but the angles ÀCD, ACB are equal (I. 13.) to two right angles ; therefore also the angles CBA, BAC, ACB are equal to two right angles. Wherefore, if a side, &c. Cor. 1. All the interior angles of any rectilineal figure, together with four right angles, are equal to twice as many right angles as the figure has sides. rectilineal figure ABCDE can be divid. ed into as many triangles as the figure has sides, by drawing straight lines from a point F within the figure to each of its angles. And, by the preceding proposition, all the angles of these triangles are equal to twice as many right angles as there are triangles, that is, as there are sides of the figure ; and the same angles are equal to the angles of the figure, together with the angles at the point F, which is the common vertex of the triangles ; that is, (I. 13. cor. 2.) together with four right angles. Therefore all the angles of the figure, together with four right angles, are equal (I. ax. 1.) to twice as many right angles as the figure has sides. * Another proof of this corollary may be obtained by dividing the figure into triangles by lines drawn from any angle to all the remote angles. Then each of the two extreme triangles has two sides of the polygon for two of its sides, D E For any A B A D B Cor. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. Because each interior angle ABC, and the adjacent exterior ABD, are together equal (I. 13.) to two right angles; therefore all the interior, together with all the exterior angles of the figure, are equal to twice as many right angles as there are sides of the figure; that is, by the foregoing corollary, they are equal to all the interior angles of the figure, together with four right angles; therefore all the exterior angles are equal (1. ax. 3.) to four right angles.* Cor. 3. If a triangle have a right angle, the remaining angles are together equal to a right angle; and if one angle of a triangle be equal to the other two, it is a right angle. Cor. 4. The angles at the base of a right-angled isosceles triangle are each half a right angle. Cor. 5. If two angles of one triangle be equal to two angles of another, their remaining angles are equal. Cor. 6. Each angle of an equilateral triangle is one third of two right angles, or two thirds of one right angle. Cor. 7. Hence, a right angle may be trisected by describing an equilateral triangle on one of the lines containing the right angle.t while each of the other triangles has only one side of the figure for one of its sides; and hence the number of triangles is less by two than the number of the sides of the figure. But the interior angles of the figure are evidently equal to all the interior angles of all the triangles, that is, to twice as many right angles as there are triangles, or twice as many right angles as the figure has sides, wanting the angles of two triangles, that is, four right angles. Hence, in any equiangular figure, the number of the sides being known, the magnitude of each angle compared with a right angle can be determined. Thus, in a regular pentagon, the amount of all the angles being twice five right angles wanting four, that is six right angles, each angle will be one fifth part of six right angles, or one right angle and one fifth. In a similar manner it would appear, that in the regular hexagon each angle is a sixth part of eight right angles, or a right angle and a third ; that in the regular heptagon each is a right angle aud three-sevenths; in the regular octagon a right angle and a half, &c. • It is to be observed, that if angles be taken in the ordinary meaning as understood by Euclid, this corollary and the foregoing are not applicable when the figures have re-entrant angles, that is, such as open outward. The second corollary will hold, however, if the difference between each re-entrant angle and two right angles be taken froń the sum of the other exterior angles : and the former will be applicable, if, instead of the angle which opens externally, the difference between it and four right angles be used. Both corollaries, indeed, will hold without change, if the re-entrant angle be regarded as internal and greater than two right angles ; and if to find the exterior angles the interior be taken, in the algebraic sense, from two right angles, as in this case, the re-entering angles will give negative or subtractive results. + By this principle also, in connexion with the 9th proposition, we may trisect any angle, which is obtained by the successive bisection of a right angle, such as the half, the fourth, the eighth, of a right angle, and so on. D PROP. XXXIIJ. THEOR. А B The straight lines which join the extremities of two equal and parallel straight lines, towards the same parts, are also themselves equal and parallel. Let AB, CD be equal and parallel straight lines, and joined towards the same parts by the straight lines AC, BD; AC, BD are also equal and parallel. Join (1. post. 1.) BC. Then, because AB is parallel to CD, and BC meets them, the alternate angles ABC, BCD are (I. 29.) equal; and because AB is equal (hyp.) to CD, and BC common to the two triangles ABC, DCB, and the angle ABC is equal to the angle BCD; therefore the base AC is equal (I. 4. part 1.) to the base BD, and (I. 4. part 3.) the angle ACB to CBD; and because the straight line BC meets the two straight lines AC, BD, and makes the alternate'angles ACB, CBD equal, AC is parallel (I. 27.) to BD; and it has been shown to be equal to it. Therefore the straight lines, &c. Cor. Hence a quadrilateral which has two sides equal and parallel, is (I. def. 24.) a parallelogram. PROP. XXXIV. THEOR. The opposite sides and angles of a parallelogram are equal to one another; and (2.) the diagonal bisects it. Let ADt be a parallelogram, of which BC is a diagonal; (1.) the opposite sides and angles of the figure are equal to one another : and (2.) the diagonal BC bisects it. Because AB is parallel (I. def. 24.) to CD, and BC meets them, the alternate angles ABC, BCD are equal (I. 29.) to one another ; and because AC is parallel to BD, and BC meets them, the alternate angles ACB, CBD are likewise equal : wherefore the two triangles ABC, CBD have two angles ABC, BCA in one, equal to two angles BCD, CBD in the other; therefore (I. 32. cor. 5.) the remaining angles BAC, CDB are equal, and the side BC, which is opposite to those angles is common; therefore (I. 26. part 1.) their other sides are equal, each to each, viz., AB to CD, and AC to BD. And because the angle ABC is equal to BCD, and CBD to ACB, the whole angle ABD is equal (I. ax. 2.) to the whole angle ACD; and BAC has been shown * That is, the extremities which lie towards the “same parts” or quarter. + For brevity, a parallelogram may be named by the letters placed at two opposite angles, if no ambiguity arise from so naming it. The same also may be done occasionally in regard to other rectilineal figures. G C F P Q B M N K D L to be equal to BDC; therefore the opposite sides and angles of a parallelogram are equal to one another. Also (I. 26. part 3.) the triangle ABC is equal to the triangle DCB ; that is, the diagonal BC bisects the parallelogram. Therefore the opposite sides, &c. Cor. 1. If a parallelogram have one right angle, all its angles are right angles. For since (I. 29.) the two angles BAC, ACD are together equal to two right angles, if one of them be a right angle, the other must also be a right angle ; and (by this proposition) the opposite angles are equal. Cor. 2. Hence we can divide a given straight line AB into any proposed number of equal parts. To do this, draw AC making any angle with AB, and draw (1. 31.) BD parallel to AC. In AC take any point E, and make (I. 3.) EF, FG, A BH, HK, KL, each equal to AE; the number of segments on each of the parallels being less by one than the number of the parts into which AB is to be divided. Draw (I. post. 2.) the straight lines EL, FK, GH, cutting AB in M, N, O: AM, MN, NO, OB are the parts required. Draw MP parallel to AC ; then (1. 33.) EL, FK, GH are parallel ; and AE and MP are equal, each of them being equal (const. and I. 34.) to EF. But (I. 29.) the angles EAM, AME are respectively equal to PMN, MNP; and therefore (1. 26.) AM is equal to MN. In the same manner, BO may be proved to be equal to ON: and by drawing NQ parallel to AC, MN may be proved in a similar manner to be equal to NO.* Cor. 3. If two parallelograms bave an angle of the one equal to an angle of the other, and the sides containing the equal angles respectively equal, the parallelograms are equal to one another. This will be shown by drawing diagonals subtending the equal angles; as (I. 4.) the triangles thus formed are equal ; and, by this proposition, the parallelograms are respectively double of those triangles. Cor. 4. If two straight lines make an angle, two others parallel to them contain an equal or a supplemental angle. Thus, A and the vertical angle made by producing BA, CA, through A, are each equal to D; while the angle made by BA and the continuation of CA, or by CA and the continuation of BA, is equal to the supplement (I. def. 38.) of D. The construction might be effected as in the 9th of the sixth book without the parallel BD. The method here given, however, is easier in practice, from the facility with which the parallels EL, FK, &c., are drawu. This problem might be considered unnecessary here, as it is the same in substance as the 9th of the sixth book. Besides the easier construction here given, however, it is so useful, that the student should be early acquainted with it. |