THEOREM 23. If a straight line intersects two other straight lines and makes either a pair of alternate angles equal, or a pair of corresponding angles equal, or a pair of interior angles on the same side supplementary; then, in each case, the two pairs of alternate angles are equal, and the four pairs of corresponding angles are equal, and the two pairs of interior angles on the same side are supplementary. Part. En. Let the straight line ABCD intersect the two straight lines EF, GH, and make the alternate angles EBC, BCH equal; then will the other alternate angles FBC, BCG be equal, and the four pairs of corresponding angles be equal, and the two interior angles on the same side be supplementary. Because . EBC=BCH, (Hyp.) and EBC= ABF being vertically opposite angles, (Th. 4.) therefore ABF=BCH; therefore also their supplements, the angles ABE and BCG are equal. Therefore also the angles which are respectively vertically opposite to these angles are equal, that is, the angle CBF=DCH, and EBC=GCD. Again, because the angle EBC= the alternate angle BCH add to each the angle CBF; therefore the two angles EBC, CBF are equal to the two CBF, BCH; but the two EBC, CBF are together equal to two right angles ; therefore the two CBF, BCH are together equal to two right angles. And in the same way it may be shewn that if two corresponding angles are given equal, or if two interior angles on the same side are supplementary, then the alternate angles will be equal. Cor. Hence if two parallel straight lines are intersected by a third straight line, the corresponding angles are equal, and the interior angles on the same side are supplementary; and conversely. THEOREM 24. Straight lines which are parallel to the same straight line are parallel to one another*. Part. En. Let A and B be each of them parallel to X, it is required to prove that A is parallel to B. Proof. If A intersected B, then two intersecting lines, A, B would each be parallel to a third line X, which is impossible, by Axiom 5. Therefore A does not intersect B, that is, A is parallel to B. Q. E. D. * Euclid, 1. 30. REMARKS. It must be observed that two parallels, and a straight line intersecting them, are a special case of the triangle, the vertex, or intersection of two of the lines, being removed to an infinite distance. In Th. 9 it was proved that the exterior angle of a triangle is greater than the interior and opposite angle: from which the contra-positive theorem (Th. 21) logically follows, that if the exterior angle is equal to the interior and opposite angle, the lines do not form a triangle. Theorem 22 is in fact proved by the rule of identity (p. 4). Since there is only one straight line through B that makes the alternate angles equal ; and only one straight line through B that is parallel to GH; (Ax. 5.) and the line that makes the alternate angles equal is the parallel ; (Th. 21.) therefore the parallel makes the alternate angles equal. THEOREM 25. If one side of a triangle be produced the exterior angle will be equal to the two interior and opposite angles, and the three interior angles of a triangle are together equal to two right angles. Let one side BC of the triangle ABC be produced to D: then shall the angle ACD= the sum of the angles ABC, CAB; and the three angles ABC, BCA, CAB shall be together equal to two right angles. Proof. For if through C a line CH were drawn parallel to BA, the angle HCD= the corresponding angle ABC, (Th. 22.) and the angle ACH= the alternate angle BAC; (Th. 22.) .:: the whole angle ACD= the two angles ABC+ BAC. Again, if ACB be added to these, the two angles ACD + ACB = the three angles ABC + BCA + CAB. But ACD+ ACB = two right angles; (Th. 4.) therefore ABC + BCA + CAB = two right angles *. Q. E. D. Cor. In a right-angled triangle the two acute angles together make up one right angle. THEOREM 26. The interior angles of any polygon are together less than twice as many right angles as the figure has sides by four right angles. Part. En. Let ABCDE be any polygon; it is required to prove that its interior angles are together less than twice as many right angles as the figure has sides by four right angles. Proof. Take any point o within the polygon; and join OA, OB, OC, OD, OE. Then there are as many triangles having O as a common vertex as the figure has sides. And, since the interior angles of a triangle are equal to two right angles; (Th. 25.) * Euclid, 1. 32. therefore all the angles of all the triangles are equal to twice as many right angles as the figure has sides. : But all the angles of all the triangles make up all the angles of the polygon together with the angles at 0, which are equal to four right angles ; (Th..4. Cor.) therefore all the angles of the polygon, together with four right angles, are equal to twice as many right angles as the figure has sides; that is, all the angles of the polygon are together less than twice as many right angles as the figure has sides by four right angles*. Cor. The exterior angles of any convex polygon are together equal to four right angles. Let ABCDE be a convex polygon having all its sides AB, BC, CD, DE, EA produced; it is required to prove that the sum of its exterior angles is equal to four right angles. Proof. Each interior angle together with its adjacent exterior angle are equal to two right angles; (Th. 2.) therefore all the interior angles together with all the exterior angles are equal to twice as many right angles as the figure has sides; * Euclid, 1. 32. Cor. I. |