1. The four angles AGF, CHE, BGF, and DHE, situated within the parallels, are called interior angles. 2. The four angles AGE, CHF, BGE, and DHF, situated without the parallels, are called exterior angles. WHEN COMPARED TWO AND TWO: 1. The two angles AGF and CHE, as well as BGF and DHE, are called interior angles on the same side; that is, on the same side of the secant or cutting-line. 2. The two angles AGE and CHF, as well as BGE and DHF, are called exterior angles on the same side. 3. The two angles AGF and DHE, as well as CHE and BGH, being interior angles on different sides of the secant, are called alternate interior angles. 4. The two angles AGE and DHF, as well as CHF and BGE, are called alternate exterior angles. We have noticed two pairs of each of the four distinct kinds of angles. 5. There are still four pairs of angles, referred to as Corresponding angles, namely: AGE and CHE, AGF and CHF, BGE and DHE, BGH and DHF. These last angles, which establish the direction of the lines, are sometimes called opposite exterior and interior angles. THEOREM XVII. If two straight lines are cut by a third line, making the sum of the two interior angles on the same side equal to two right angles, the two lines will be parallel. If the two lines AB and CD are cut by the line EF, making BGH+DHG = 2 right angles. these lines will be parallel. Cor. If two straight lines are cut by a third line, they will be parallel: I. When the sum of the exterior angles on the same side is equal to two right angles. II. When the alternate interior angles are equal. We have EGB= AGH, and FHD = CHG (T. I.), hence AGH+CHG=2 right angles, which agrees with the Theorem itself, hence AB and CD are parallel. Secondly. Suppose AGH = GHD. We have AGH = EGB (T. I.), consequently EGB=GHD; that is, the lines AB and CD have the same direction, and are therefore parallel. Thirdly. Suppose EGB = CHF. We have CHF = GHD, consequently EGB=GHD, and the lines have the same direction, and are therefore parallel. Fourthly. Since the corresponding angles are equal, the lines have the same direction, and are therefore parallel. THEOREM XVIII. If two parallels are cut by a third line, the sum of the two interior angles on the same side is equal to two right angles. Since the lines are parallel they have the same direction, and their corresponding angles must be equal. We therefore have EGB=GHD; to each add BGH, and we shall have EGB+BGH=BGH+GHD. A C F G E B H D But EGB+BGH = 2 right angles (T. I.), consequently BGH +GHD=2 right angles. Cor. It follows immediately from D. X., in connection with T. I., C. I., that if either one of the eight angles formed by the intersection of a straight line with two parallels is a right angle, the other seven will also be right. When these angles are not right, there will be four equal acute angles, and four equal obtuse angles. These angles will be supplementary. Any two of the acute angles, or any two of the obtuse angles taken together, constitute a pair of corresponding angles, and fix the direction of the parallels. Hence it readily follows, that— If two parallels are cut by a third line, we shall have: I. The sum of the exterior angles on the same side equal two right angles. II. The alternate interior angles equal. THEOREM XIX. If two straight lines are cut by a third line, and the sum of the interior angles on the same side is not equal to two right angles, the two lines will meet, if sufficiently produced. If we suppose the angle AGH+ CHG > 2 right angles, then, obviously, will the angle BGH+DHG <2 right angles, since these angles are respectively supplementary angles. This being A C -D H B supposed, these lines will meet. For if they do not meet they are parallel; but they are not parallel, since the sum of the interior angles on the same side is not equal to two right angles. Hence these lines will meet if produced sufficiently far. Scholium. It is evident they will meet in the direction of B and D, on that side of the secant line which has the sum of the interior angles less than two right angles. OF TRIANGLES. THEOREM XX. If two triangles have two sides and the included angle of the one, equal to the two sides and the included angle of the other, the triangles will be identical, or equal in all respects. In the two triangles ABC, DFG, if the side CA be equal to the side DG, and the side CB equal to the side GF, and the angle C equal to the angle G, then will the two triangles be identical, or equal in all respects. For, conceive the triangle ABC to be placed upon the triangle DFG in such a manner that the point C may coincide with the point G, and the side CA with the equal side GD. Then, since the angle G is equal to the angle C, the side CB will take the direction of the side GF. Also CB being equal to GF, the point B will coincide with the point F; consequently the side AB will coincide with DF. Therefore the two triangles are identical, and have all their other corresponding parts equal (A. IX.), namely, the side AB equal to the side DF, the angle A equal to the angle D, and the angle B equal to the angle F. Cor. Two right-angled triangles are equal, when the two sides containing the right angle of the one are respectively equal to the two sides containing the right angle of the other. THEOREM XXI. If two triangles have two sides of the one respectively equal to two sides of the other, and the included angle of the first greater than the included angle of the second, the third side of the first will be greater than the third side of the second. In the two triangles ABC and DEF, suppose CA and CB of the first to be respectively equal to FD and FE of the sec- A L ond, and the included an C F gle ACB > DFE, then will AB > DE. Apply the triangle DEF upon the triangle ABC, making EF coincide with BC, so that the triangle may, in its new position, be represented by GBC. Draw CL bisecting the angle ACG (Post. VI.), and meeting AB at L, draw GL. The two triangles ACL, GCL, are equal (T. XX.), and LG = LA. Therefore, AB = AL+LB = LG + B D A C F E F LB, which is greater than BG (T. VII.). That is, AB>DE. Although the triangle DEF, when applied to ABC, may have three different positions, as here represented, still our demonstration is alike applicable to either case. A LG B D E Cor. Conversely. If two triangles have two sides of the one respectively equal to two sides of the other, and the third side of the first greater than the third side of the second, the included angle of the first will be greater than the included angle of the second. For, if the included angle of the first is not greater than the included angle of the second, it must be either equal or less. If it were equal, the third sides would be equal (T. XX.). If it were less, the third side of the first would be less than the third side of the second, by the Theorem itself. Both these results are contrary to the hypothesis. Consequently, the included angle of the first is greater than the included angle of the second. THEOREM XXII. If two triangles have two angles and the interjacent side of the one equal to two angles and the interjacent side of the other, the triangles will be identical, or equal in all respects. |