Scholium. When two parallel straight lines AB, CD, are met by a third line FE, the angles which are formed take particular names. Interior angles on the same side, are those which lie within the parallels, and on the same side of the secant line: thus, OGB, GOD, are interior angles on the same side; and so also are the the angles OGA, GOC. I Let the parallels AB, CD,be E Alternate angles lie within the parallels, and on different sides of the secant line: AGO, DOG, are alternate angles; and so also are the angles COG, BGO. For, if OGB+ GOD be not *Alternate exterior angles lie without the parallels, and on different sides of the secant line: EGB, COF, are alternate exterior angles; so also, are the angles AGE, FOD. Opposite exterior and interior angles lie on the same side of the secant line, the one without and the other within the parallels, but not adjacent: thus, EGB, GOD, are opposite exterior and interior angles; and so also, are the angles AGE, GOC. F Cor. 1. If a straight line EF, meet two straight lines CD, AB, making the alternate angles AGO, GOD, equal to each other, the two lines will be parallel. For, to each add the angle OGB; we shall then have, AGO+OGB=GOD+OGB; but AGO+OGB is equal to two right angles (Prop. I.); hence GOD+OGB is equal to two right angles: therefore, CD, AB, are parallel. Cor. 2. If a straight line EF, meet two straight lines CD, AB, making the exterior angle EGB equal to the interior and opposite angle GOD, the two lines will be parallel. For, to each add the angle OGB: we shall then have EGB+OGB=GOD +OGB: but EGB+OGB is equal to two right angles; hence, GOD+OGB is equal to two right angles; therefore, CD, AB, are parallel. B D PROPOSITION XX. THEOREM. If a straight line meet two parallel straight lines, the sum of the interior angles on the same side will be equal to two right angles. E B H D right angles; then III and CD will be parallel (Prop. XIX.), and hence we shall have two lines GB, GH, drawn through the same point G and parallel to CD, which is impossible (Ax. 12.): hence, GB and GH should coincide, and OGB+GOD is equal to two right angles. In the same manner it may be proved that OGA+GOC is equal to two right angles. Cor. 1. If OGB is a right angle, GOD will be a right angle also therefore, every straight line perpendicular to one of two parallels, is perpendicular to the other. Cor. 2. If a straight line meet two parallel lines, the alternate angles will be equal. A E T Let AB, CD, be the parallels, and FE the secant line. The sum OGB+ GOD is equal to two right angles. But the sum OGB+OGA is also equal to two right angles (Prop. I.). Taking from each, the angle OGB, and there remains OGA=GOD. In the same manner we may prove that GOC=OGB. B D Cor. 3. If a straight line meet two parallel lines, the opposite exterior and interior angles will be equal. For, the sum OGB+GOD is equal to two right angles. But the sum OGB +EGB is also equal to two right angles. Taking from each the angle OGB, and there remains GOD=EGB. In the same manner we may prove that AGE=GOC. Cor. 4. We see that of the eight angles formed by a line cutting two parallel lines obliquely, the four acute angles are equal to each other, and so also are the four obtuse angles. E PROPOSITION XXI. THEOREM. If a straight line meet two other straight lines, making the sum of the interior angles on the same side less than two right angles, the two lines will meet if sufficiently produced. I Let the line EF meet the two lines CD, IH, making the sum of the interior angles OGII, A GOD, less than two right anglcs: then will IH and CD meet if sufficiently produced. C F For, if they do not meet they are parallel (Def.12.). But they are not parallel, for if they were, the sum of the interior angles OGH, GOD, would be equal to two right angles (Prop. XX.), whereas it is less by hypothesis: hence, the lines IH, CD, are not parallel, and will therefore meet if sufficiently produced. H B D Cor. It is evident that the two lines IH, CD, will meet on that side of EF on which the sum of the two angles OGH, GOD, is less than two right angles PROPOSITION XXII. THEOREM. Two straight lines which are parallel to a third line, are parallel to each other. Let CD and AB be parallel to the third line EF; then are they parallel to each other. Draw PQR perpendicular to EF, and cutting AB, CD. Since AB is parallel to EF, PR will be perpendicular to AB (Prop. E XX. Cor. 1.); and since CD is parallel to EF, PR will for a like reason be perpen-C dicular to CD. Hence AB and CD are. perpendicular to the same straight line; hence they are parallel (Prop. XVIII.). A PROPOSITION XXIII. THEOREM. Two parallels are every where equally distant. Two parallels AB, CD, being CH given, if through two points E and F, assumed at pleasure, the straight lines EG, FH, be drawn perpendicular to AB,these straight A lines will at the same time be perpendicular to CD (Prop. XX. Cor. 1.): and we are now to show that they will be equal to each othe T E If GF be drawn, the angles GFE, FGH, considered in reference to the parallels AB, CD, will be alternate angles, and therefore equal to each other (Prop. XX. Cor. 2.). Also, the straight lines EG, FH, being perpendicular to the same straight line AB, are parallel (Prop. XVIII.); and the angles EGF, GFH, considered in reference to the parallels EG, FH, will be alternate angles, and therefore equal. Hence the two triangles EFG, FGH, have a common side, and two adjacent angles in each equal; hence these triangles are equal (Prop. VI.); therefore, the side EG, which measures the distance of the parallels AB and CD at the point E, is equal to the side FH. which measures the distance of the same parallels at the point F. G D B PROPOSITION XXIV. THEOREM. If two angles have their sides parallel and lying in the same direction, the two angles will be equal. Let BAC and DEF be the two angles, having AB parallel to ED, and AC to EF; then will the angles be equal. B A PROPOSITION XXV. THEOREM. E For, produce DE, if necessary, till it meets AC in G. Then, since EF is parallel to GC, the angle DEF is equal to DGC (Prop. XX. Cor. 3.); and since DG is parallel to AB, the angle DGC is equal to BAC; hence, the angle DEF is equal to BAC (Ax. 1.). E G F Scholium. The restriction of this proposition to the case where the side EF lies in the same direction with AC, and ED in the same direction with AB, is necessary, because if FE were produced towards II, the angle DEH would have its sides parallel to those of the angle BAC, but would not be equal to it. In that case, DEH and BAC would be together equal to two right angles. For, DEH+DEF is equal to two right angles (Prop. I.); but DEF' is equal to BAC: hence, DEH +BAC is equal to two right angles. F B In every triangle the sum of the three angles is equal to two right angles. Let ABC be any triangle: then will the angle C+A+B be equal to two right angles. A D For, produce the side CA towards D, and at the point A, draw AE parallel to BC. Then, since AE, CB, are parallel, and CAD cuts them, the exterior angle DAE will be equal to its inte-C rior opposite one ACB (Prop. XX. Cor. 3.); in like manner, since AE, CB, are parallel, and AB cuts them, the alternate angles ABC, BAE, will be equal: hence the three angles of the triangle ABC make up the same sum as the three angles CAB, BÃE, EAD; hence, the sum of the three angles is equal to two right angles (Prop. I.). E Cor. 1. Two angles of a triangle being given, or merely their sum, the third will be found by subtracting that sum from two right angles. C* Cor. 2. If two angles of one triangle a re respectively equal to two angles of another, the third angles will also be equal, and the two triangles will be mutually equiangular. Cor. 3. In any triangle there can be but one right angle : for if there were two, the third angle must be nothing. Still less, can a triangle have more than one obtuse angle. Cor. 4. In every right angled triangle, the sum of the two acute angles is equal to one right angle. Cor. 5. Since every equilateral triangle is also equiangular (Prop. XI. Cor.), each of its angles will be equal to the third part of two right angles; so that, if the right angle is expressed by unity, the angle of an equilateral triangle will be expressed by. Cor. 6. In every triangle ABC, the exterior angle BAD is equal to the sum of the two interior opposite angles B and C. For, AE being parallel to BC, the part BAE is equal to the angle B, and the other part DAE is equal to the angle C. PROPOSITION XXVI. THEOREM. The sum of all the interior angles of a polygon, is equal to two right angles, taken as many times less two, as the figure has sides. Let ABCDEFG be the proposed polygon. If from the vertex of any one angle A, diagonals p, AC, AD, AE, AF, be drawn to the vertices of all the opposite angles, it is plain that the polygon will be divided into five triangles, if it has seven sides; into six triangles, if it has eight; and, in general, into as many triangles, less two, as the polygon has sides; for, these triangles may be considered as having the point A for a common vertex, and for bases, the several sides of the polygon, excepting the two sides which form the angle A. It is evident, also, that the sum of all the angles in these triangles does not differ from the sum of all the angles in the polygon: hence the sum of all the angles of the polygon is equal to two right angles, taken as many times as there are triangles in the figure; in other words, as there are units in the number of sides diminished by two. Cor. 1. The sum of the angles in a quadrilateral is equal to two right angles multiplied by 4-2, which amounts to four |