Proof. For BA and AC being equal to ED, DF, if the angle A were unequal to the angle D, then the base BC would be unequal to the base EF (by Theorem 17). But BC is equal to EF; And therefore the angle A= the angle D; and then (by Theorem 16), the angles B and C are equal to the angles E and F. THEOREM 19. Converse of the Opposite. If two triangles have two sides of the one equal to two sides of the other but the bases are unequal, the angles included by the equal sides shall be unequal, the greater base having the greater angle opposite to it. = Let ABC, DEF be two triangles in which BA, AC DE, EF, but BC is > DF Then will the angle A be> the angle E. B Proof. For if the angle A were equal to the angle E, then the base BC would be equal to the base DF (Th. 16); which it is not. Therefore the angle A is not equal to the angle E. Nor is the angle A less than the angle E; For then the base BC would be less than the base DF (Th. 17); but it is not. Therefore the angle A is greater than the angle E. #12.2, but let the angle R.4C be greater than the angle 12. then shall the base 'be greater than the base DF. 2. For by the triangle FFD be Plood an that falls on 43, but EF Follow the aude of 45 opposite to F But BC is equal to EF; And therefore the angle A = the angle D; and then (by Theorem 16), the angles B and C are equal to the angles E and F. THEOREM 19. Converse of the Opposite. If two triangles have two sides of the one equal to two sides of the other but the bases are unequal, the angles included by the equal sides shall be unequal, the greater base having the greater angle opposite to it. Let ABC, DEF be two triangles in which BA, AC =DE, EF, but BC is > DF Then will the angle A be> the angle E. F the angle A were equal to the angle E, ld be equal to the base DF (Th. 16); re the angle A is not equal to the s than the angle E; C would be less than the base DF A is greater than the angle E. THEOREM 20. If two triangles have two sides of the one equal to two sides of the other, and the angle opposite that which is not the less of the two sides of the one equal to the corresponding angle of the other, the triangles shall be equal in all respects. Let ABC, DEF be the two triangles, having the sides BA, AC equal to the sides ED, DF respectively, of which AC is not less than AB, and having also the angle B = the angle E. Then shall the triangles be equal in all respects. = Proof. For if AC=AB, and DE DF, the angles B, C, E, Fare equal, and the remaining angle A= the remaining angle D, and the triangles are equal in all respects by Theorems (15) or (16). AA If AC is > AB, then since AB = DE it could be placed so as to coincide with it; and since the angle B = the angle E, the line BC would coincide in direction with EF, and the point C would fall somewhere on EF, or on EF produced through F. It remains to see the effect of the only remaining condition that AC= DF. Now the line AC has one extremity A on D, and the other extremity C somewhere on EF, or on EF produced through F, and AC=DF. But there is no oblique equal to DF, and not coinciding with it, that can be drawn from D to any point in EF; for since DF is > DE it is more remote from the perpendicular (Th. 12, Cor.); and therefore the oblique equal to DF would lie on the other side of DE, and have its extremity in EF produced through E. Hence AC would coincide with DF, and the triangles are therefore equal in all respects. COR. I. If the side opposite the given angle were less than the side adjacent, there would be two triangles, as in the figure; and the proof given above is inapplicable. This is called the ambiguous case. COR. 2. If the given angle is a right angle, the side opposite to it must be greater than the side adjacent; by Th. II. Hence if two right-angled triangles have the hypothenuse and one side of the one equal respectively to the hypothenuse and one side of the other, the triangles are equal in all respects. This corollary is of very frequent use. COR. 3. A similar property is obviously true of two obtuse-angled triangles. |