The two angles CBA, ABD equal the three -angles CBE, EBA, ABD (d). The two angles DBE, EBC equal the three angles CBE, EBA, ABD (c). (e) By Axiom I the two angles CBA, A B D equal the two angles D BE, E BC. The angles DBE, E B C are two right angles. The angles CBA, ABD equal the angles DBE, EBC. .. The angles CBA, ABD are together equal to two right angles. Q. E. D. EXERCISE.-Prove that the five angles, ACD, DCE, ECF, FC G, G C B, are together equal to two right angles. (Draw a line at right angles to A B). THEOREM (Euclid I. 14). Repeat. The enunciation of Euc. I. 13 and Axioms 3a and 11. General Enunciation. If, at a point in a straight line, two other straight lines, on the opposite sides of it, make the adjacent angles together equal to two right angles, those two straight lines shall be in one and the same straight line. B To prove that those two straight lines are in one and the same straight line. Particular Enunciation. Given. (a) The straight line A B. (b) The point B in A B. D (c) Two straight lines at the point B, on opposite sides of A B. (d) The angles ABC, ABD equal to two right angles. Required. To prove that CB is in the same straight line with BD. Proof (Indirect)— Let us suppose that CB is not in the same straight line with B D. Then draw BE, and (a) assume that BE and CB are in the same straight line. A B E C The angles which one straight line makes with another on one side of it, are equal to two right angles (Euc. I. 13). CBE is assumed to be a straight line (a), and the line AB makes with it on the same side, the angles A B C, A BE. (b) .. The angles ABC, ABE are together equal to two right angles. The angles ABC, A B D were given (d) equal to two right angles. The angles ABC, ABD have been proved (b) equal to two right angles. .. The angles ABC, ABD are equal to the angles ABC, ABE. (Axiom 11.) From these equals take away the common angle ABC. (The angle ABC being found in both pairs of angles, is called the common angle.) Then the remaining angle ABD is equal to the remaining angle A B E (Axiom 3a). That is to say, we have proved the greater angle ABD equal to the lesser angle A B E. As this is absurd and impossible, it is also absurd to suppose, as we did (a), that the line BE can be in the same straight line with CB. THEOREM (Euclid I. 16). Repeat. The enunciations of Euc. I. 15, I. 4, and Axiom 9. General Enunciation. If one side of a triangle be produced, the exterior angle shall be greater than either of the interior opposite angles. Particular Enunciation. Given. The tri angle ABC (Fig. 1). with the side B C produced to D. Required. To prove that the angle ACD is greater than either of the angles ABC or BAC. Fig:1 B с D Of the four angles in this figure, A CD is called the exterior angle, being outside the triangle; the other three angles are called interior angles, being inside the triangle. Of the three interior angles, A CB is called the interior adjacent angle, being next to the exterior angle; ABC and BAC are called the interior opposite angles, being opposite to the exterior angle. Construction (for purposes of proof, Fig. 2). Bisect A C, and call the point of intersection E. Produce the line BE to F, making EF equal to BE. Ist Proof. (We have to prove that angle A CD is greater than either of the angles ABC or BAC.) AC was bisected. (a).. AE is equal to E C (construction). And (6) E F was made equal to B E (const.). The lines AC and BF cut one another (Euc. I. 15). Fig. 2 B (c) .. The angle A E B is equal to the angle CEF. have two sides of the equal to two sides of and also the angles in- The remaining an- D We use conclusions (a), (b), and (c) in applying Euc. I. 4 to the two triangles AEB, FEC. Thus: Euclid proves (I. 4) that | We have (in Fig. 2) if two triangles The two triangles AEB, FEC, having the two sides AE, EB, equal to the two sides each to each (a and b), (d) The remaining angle BAE is equal to the remaining angle But the whole angle A CD is greater than its part ECF |