Proof.- BCG can be shown, in the same manner as above, to be greater than ▲ ABC. < ACD = BCG (I. 15), and .. is greater than ABC. Prop. 17. A simple theorem follows this somewhat intricate demonstration: 'Any two angles of a triangle are together less than two right angles.' To show this we must compare them with two angles which either are or are together equal to two right angles. B A C D These are easily obtained, as you will see from a glance at the last figure, by producing one side of the triangle; then the exterior angle, together with the adjacent interior angle, are equal to two right angles (I. 13). Produce BC, the base of triangle ABC, to D, then angles ACD, ACB are together equal to two right angles. Now we know that angle ACD is greater than either of the interior opposite angles; it follows, then, that angle ACB and either of these (interior and opposite) angles must be together less than two right angles. But are these two opposite angles, taken together, less than two right angles? If we produce another side, e.g. AB to E, we shall find that the same proof applies, and that they are also less than two right angles. SUMMARY OF PROPOSITION XVII., THEOREM 10. Any two angles of a triangle are together less than two right angles. Cons.-Produce one side of . Proof. The ext. ▲ and adjacent int. Zare greater than adjacent int. and either of opposite int. s (I. 16), and -D = two rt. s (I. 13).. either of these pairs of int.s are less than two rt. s. B C Produce another side. Repeat proof for the remaining pair of angles. We will conclude this lesson with two short theorems, pointing out another relation between the sides and angles of a triangle. These are the converse of each other. (a) The greater side of every triangle has the greater angle opposite to it (I. 18). (B) The greater angle of every triangle has the greater side opposite to it. Let ABC be any triangle whose side AC is greater than AB. We have to show that angle ABC is greater than angle ACB. What do we already know about the relation of sides to angles in the same triangle? From the 5th and 6th propositions we learn that equal angles are opposite equal sides. Let us compare an isosceles triangle with the one given. Since AC is greater than AB, cut off AD = AB (I. 3). Join BD, then ABD is an isosceles triangle. Can we observe any relation between the equal angles of the isosceles triangle and those which we are required to com pare? First, each of these equal angles D Has either of them any relation to angle ACB? Yes; angle ADB is the exterior angle of the triangle DBC, and is therefore greater than the interior and opposite angle ACB. Then angle ADB is less than angle ABC and greater than angle ACB, 'much more then must angle ABC be greater than angle ACB.' E SUMMARY OF PROPOSITION XVIII., THEOREM 11. The greater side of every triangle has the greater angle B Z ACB (I. 16). Cons.-Cut off from greater side a part equal to less. Join DB. Proof. ADB = ▲ ABD (I. 5), is less than ABC. < ADB is greater than int. and opposite . ABC is greater than ACB. I. 19. The converse theorem, as is often the case, is proved by the indirect method. Angle ABC being given greater B We know that than angle ACB, we have to show that side AC is greater than the side AB. If AC is not greater than AB, it must either be equal to it or less. What would be the consequence if AB= AC? ABC would be equal to ▲ ACB (I. 5), but it is not (being given greater). What would be the consequence if AC is less than AB? We know that the angle ACB would be greater than the angle ABC (I. 18), but it is not. It remains, then, that AC must be greater than AB. SUMMARY OF PROPOSITION XIX., THEOREM 12. The greater angle of every triangle has the greater side opposite. Cons.-Nil. B Proof (Indirect).-AB is not equal to AC (I. 5). AB is not greater than AC (I. 18), .. it is less. LECTURE IX. PROPOSITIONS 20-21. I. 20. WE come to the well-known and much-derided 20th Proposition, said by the Epicureans to be 'evident even to asses.' 'Any two sides of a triangle are greater than the third.' Though this fact seems very obvious, its mathematical proof is not easy, besides which, this and other plain propositions ought to be proved in order to avoid increasing the number of axioms. Let ABC be any triangle, we will endeavour to show that sides BA, AC are together greater than BC. In order to compare these lines readily, make one line equal to the two BA, AC, by producing BA to D, making AD equal to AC (I. 3). Then BD will represent the lines BA, AC, and we have to show that BD is greater than BC. In the last proposition one line was shown to be greater than another because it was opposite a greater angle of a triangle, BD, BC will be sides of a triangle if we join DC. Let us examine the angles in the figure thus obtained, and see if we can arrive at this proof. B Is angle BCD greater than angle BDC ? = AC. BCD is greater than ▲ ADC (BDC). .. BD is greater than BC, and BD = BA and AC. We can repeat this proof to show that BC, CA are greater than BA, and AB, BC greater than AC, and then the theorem is fully demonstrated. SUMMARY OF PROPOSITION XX., THEOREM 13. Any two sides of a triangle are together greater than the third. D Produce BA to D, make AD = AC. Join DC. < ACD = 2 ADC (I. 5). BCD is greater than ▲ ACD, and c consequently than ADC. .. BD (= BA, AC) is greater than BC (I. 19). I. 21. We pass to Prop. XXI., the last of this group of theorems. If from the ends of the side of a triangle there be drawn two straight lines to a point within it, these will be less than the other sides of the triangle, but will contain a greater angle. Let ABC be any triangle, and from B, C, ends of side BC, draw straight lines BD, CD to a point D within the triangle, BD and DC shall be less than BA and AC, but the angle BDC shall be greater than the angle ВАС. B D E C I. To show that BD, DC are less than BA, AC. In order to connect these two pairs of lines, produce one of the lines. meeting at D to meet the opposite side of the triangle BAC in E. We have now BA + AC = BA + AE + EC (A). Let us compare these groups of lines, beginning at the bottom of the chain. DC must be less than DE + EC, for any two sides of a triangle are greater than the third, and BD is common to groups B and C, therefore BD + DC are less than BE + EC. Again, BE is less than BA + AE (I. 20), and EC is |