A E G C D H Because the angle EGB is equal (hyp.) to GHD, and EGB (I. 15) to AGH, the angle AGH is equal (I. ax. 1) to GHD; and they are alternate angles: therefore AB is parallel (I. 27) to CD. Again, because the angles BGH, GHD are equal (hyp.) to two right angles; and that AGH, BGH are also equal (I. 13) to two right angles; the angles AGH, BGH are equal (I. ax. 1 and 11) to BGH, GHD; take away the common angle BGH; therefore the remaining angles AGH, GHD are equal; and they are alternate angles; therefore AB is parallel (I. 27) to CD.* Wherefore two straight lines, &c. PROP. XXIX. THEOR.-If a straight line fall upon two parallel straight lines, (1) it makes the alternate angles equal to one another; (2) the exterior angle equal to the interior and remote upon the same side; and (3) the two interior angles upon the same side together equal to two right angles.t Let the straight lines AB, CD be parallel, and let EF fall upon them; then (1) the alternate angles AGH, GHD are equal to one another; (2) the exterior angle EGB is equal to the interior and remote, upon the same side, GHD; and (3) the two interior angles BGH, GHD upon the same side are together equal to two right angles. A G C D H P For, if AGH be not equal to GHD, one of them must be greater than the other; let AGH be the greater, and to each of them add the angle BGH: therefore the angles AGH, BGH are greater (I. ax. 4) than the angles BGH, GHD; but the angles AGH, BGH are equal (I. 13) to two right angles: wherefore the angles BGH, GHD are less than two right angles. But (I. ax. 12) those straight lines, which with another straight line falling upon them, make the interior angles on the same side less than two right angles, meet together, if continually produced: therefore the straight lines AB, CD, if produced far enough, will meet: but (I. def. 11) they never meet, since (hyp.) they are parallel : therefore the angle AGH 1 *It is easy to prove, that if the exterior angles on the same side of the line which falls on the other two, be together equal to two right angles, the straight lines are parallel. This proposition will readily lead us to admit the truth of what is called the 12th axiom. It is here proved, that if the two angles BGH, GHD be together equal to two right angles, AB and CD are parallel, and consequently do not meet. Now, if through G a straight line be drawn in the angle BGH, and consequently making with GH an angle less than BGH, so that the two angles which GH makes with it and with HD are together less than two right angles, we will readily admit, that if that line and HD be continually produced, they will at length meet: which is the 12th axiom. The truth of this axiom, therefore, will be admitted by any one who has read the Elements thus far, but not, as is the case, with regard to the other axioms, by a person who is unacquainted with the principles of geometry. The first part of this proposition is the converse of the 27th, and the second and third parts are the converses of the first and second parts of the 28th. E G A is not unequal to the angle GHD, that is, it is equal to it. But D H Cor. 1. 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 a straight line drawn from G to any point H in CD, 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 a third 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. A L G B K с D H E F Let the straight line LH cut AB, CD, EF; and because LH cuts the parallel straight lines AB, EF, the angle LGB is equal (I. 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 LHF; and it has been shown that the angle LGB is equal to LHF: wherefore also LGB is equal (I. ax. 1) to LKD, the interior and remote angle on the same side of LH: therefore AB is parallel (I. 28, part. 1) to CD. Wherefore two straight lines, &c. 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. E 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. A C D B 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. 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) the 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. B A C D E 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 1) 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 has been 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, BAC, ACB; but the angles ACD, ACB are equal (I. 13) to two right angles: therefore also the angles CBA, BAC, ACB are equal to two right angles. Where fore, 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. E For any rectilineal figure ABCDE can be divided 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. I) to twice as many right angles as the figure has sides.** A B * 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, 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, Cor. 2. All the exterior angles of any rectilineal figure are together equal to four right angles. A 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 D 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 (I. ax. 3) to four right angles.* B 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. † PROP. XXXIII. THEOR.-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, which are joined towards the same parts by the straight lines AC, BD; AC, BD are also equal and parallél. Join (I. post. 1) BC. Then, because AB is parallel to CD, and 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 and 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 cach re-entrant angle and two right angles be taken from 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. That is, the extremities which lie towards the "same parts" or quarter. B 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. Also, 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 AD* 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 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: also 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. Again, 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 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, part 3) 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. A rectangle (I. def. 26), therefore, has all its angles right angles. Cor. 2. If two parallelograms have an angle of the one equal to an angle of the other, and the sides containing the equal angles 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 eceasionally with regard to other rectilineal figures. |