.. The angle ACD is greater than the angle BAC (d). Then if B C be bisected at E, and A C produced (Fig. 3), it can be proved in the same manner, as above, that the angle (e) BCG is greater than the angle ABC. But the lines BD and A G cut one another in the point C (Euc. I. 15). (f). The angle BCG is equal to the angle ACD. Hence it follows from (e) and (ƒ) that the angle A CD is greater than the angle ABC, and it has also been proved greater than the angle B AC. Q. E. D. 2nd Proof (Contracted, in the words of Euclid). Because (Fig. 2) AE is equal to E C, and BE to EF (Construction), the two sides AE, E B are equal to the two sides CE, EF each to each; and the angle A E B is equal to the angle CE F because they are vertical angles (Euc. I. 15). .. the triangle A E B is equal to the triangle CEF, and the remaining angles equal to the remaining angles, each to each, to which the equal sides are opposite (Euc. I. 4). .. the angle BAE is equal to the angle E CF. But the angle ECD is greater than the angle ECF (Axiom 9). .. The angle ACD is greater than the angle BAC. In the same manner, if B C (Fig. 3) be bisected, and the side AC be produced to G, it may be shown that the angle BCG, that is the angle ACD (Euc. I. 15), is greater than the angle ABC. Q. E. D. EXERCISES.-I. Prove, as above, that angle BCG (Fig. 3) is greater than angle ABC. II. In Fig. 3 of the last theorem, prove that angle BAC is double of the angle EFC. III. In the same figure, if A E is at right angles to B C, prove that A B and A C are equal. IV. In the same figure, if A E is at right angles to BC, prove that angle A CF is double of the angle ABC. THEOREM (Euclid I. 17). Repeat. The enunciation of Euc. I. 13, and of General Enunciation. Enun ciation. Given. The triangle B. A Required. To prove that any two of its angles taken together are less than two right angles. Construction. We will prove that angles A B C and A CB, are together less than two right angles. Produce the side B C to D. B Proof (by Euc. I. 16). The exterior angle of any triangle is greater than either of the interior opposite angles. A CD is the exterior angle of triangle A B C. (a) .. ACD is greater than the interior opposite angle ABC. ACD is greater than ABC (a). ABC, ACB are less than ACD, ACB (b). ACD, ACB are equal to two right angles (Euc. I. 13). .. ABC, ACB are less than two right angles. Q. E. D. EXERCISES.-I. Prove, by producing the side A C, that angles B A C and A CB are together less than two right angles. II. Prove, by producing the side A B, that angles ABC and B A C are together less than two right angles. III. Prove that angles A B C, and ACB in this figure are together less than two right angles, without producing either side. (Notice that A B D and ACD are two triangles, and that ADC is the exterior angle of triangle A BD; while ADB is the exterior angle of triangle A CD (Euc. I. 16). THEOREM (Euclid I. 18). Repeat.-Euc. I. 5; Euc. I. 16; and Axiom 9. General Enunciation. The greater side of every triangle has the greater angle opposite to it. B A Particular Enunciation. Given. The triangle A B C, having the side A C greater than AB. Required. To prove that angle ABC is greater than angle ACB. A Construction. From A C, the greater, cut off a part A D, equal to AB the less. Join BD. Proof. BDC is a triangle, having its side DC produced to A. Then ADB is the exterior angle of the triangle BDC. (a) The exterior angle ADB is greater than the interior opposite angle DC B (Euc. I. 16). ABD is an isosceles triangle having the side A B equal to the side A D (by construction). (b) .. angle A D B is equal to angle Á BD (Euc. I. 5). angle A D B is greater than angle DCB (a). angle ADB is equal to angle ABD (b). (c) angle ABD is greater than angle DC B, but angle ABD is part of angle A B C. .. ABC is still greater than angle ACB (c). Q E. D. EXERCISES.-I. In the figure of this proposition let the side B C be greater than the side AB. Then prove that angle B A C is greater than angle AC B. |