PROPOSITION 17.-THEOREM. Any two angles of a triangle are together less than two right angles. (References, Prop. I. 13, 16; ax. 4.) Hypothesis.-Let ABC be any triangle. Sequence.-Any two of its angles together shall be less than two right angles. Construction. Produce BC to D. Demonstration.-1. Because ACD is the exterior angle of the triangle ABC, it is greater than the interior and opposite angle ABC. (I. 16.) 2. To each of these add the angle ACB. 3. Therefore the angles ACD, ACB, are greater than the angles ABC, ACB. (ax. 4.) 4. But the angles ACD, ACB, are together equal to two right angles. (I. 13.) 5. Therefore the angles ABC, ACB, are together less than two right angles. 6. In like manner it may be proved that the angles BAC, ACB, as also the angles CAB, ABC, are together less than two right angles. Conclusion.—Therefore, any two angles, &c. Q. E. D. PROPOSITION 18. -THEOREM. The greater side of every triangle is opposite the greater angle. (References-Prop. I. 3, 5, 16.) Hypothesis.--Let ABC be a triangle, of which the side AC is greater than the side AB. Sequence.—The angle ABC shall be greater than the angle BCA. Construction. - Because AC is greater than AB, make AD equal to AB (I. 3), and join BD. Demonstration.—1. Because ADB Bis the exterior angle of the triangle BDC, it is greater than the interior and opposite angle BCD. (I. 16.) 2. But the angle ADB is equal to the angle ABD; the triangle BAD being isosceles. (Î. 5.) 3. Therefore the angle ABD is greater than the angle BCD (or ACB). 4. Much more then is the angle ABC greater than the angle ACB. Conclusion.—Therefore, the greater side, &c. Q. E. D. PROPOSITION 19.-THEOREM. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to ito (References—Prop. I. 5, 18.) Hypothesis.—Let ABC be a triangle, of which the angle ABC is greater than the angle BCA. Sequence.—The side AC shall be greater than the side AB. Demonstration.-l. If AC be not greater than AB, it must either be equal to or less than AB. 2. It is not equal, for then the angle ABC would be equal to the angle BCA (I. 5); but it is not. (hyp.) 3. Therefore AC is not equal to AB. § 4. Neither is AC less than AB, for then the angle ABC would be less than the angle BCA (I. 18); but it is not. (hyp.) 5. Therefore AC is not less than AB. 6. And it has been proved that AC is not equal to AB. 7. Therefore AC is greater than AB. Conclusion.—Therefore, the greater angle, &c. Q. E. D. PROPOSITION 20.—THEOREM. Any two sides of a triangle are together greater than the third side. (References Prop. I. 3, 5, 19; ax. 9.) Hypothesis.—Let ABC be a triangle. Sequence.-Any two sides of it are together greater than the third side. Construotion.- Produce BA to the point D, making AD equal to AC (I. 3), and join DC. Demonstration.-1. Because DA is equal to AC, the angle ADC is equal to the angle ACD. (I. 5.) 2. But the angle BČD is greater than the angle ACD. (ax. 9.) 3. Therefore the angle BCD is greater than the angle ADC (or BDC). 4. And because the angle BCD of the triangle DCB, is greater than its angle BDC, and that the greater angle is subtended by the greater side; 5. Therefore the side DB is greater than the side BC. (I. 19.) 6. But BD is equal to BA and AC. 8. In the same manner it niay be proved that AB, BC, are greater than AC, and BC, CA, greater than AB. Conclusion.—Therefore, any two sides, &c. Q. E. D. PROPOSITION 21.—THEOREM. If from the ends of the side of a triangle there be drawn tro straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shull contain a greater angle. (References—Prop. I. 16, 20; ax. 4.) Hypothesis.—Let ABC be a triangle, and from the points B, C, the ends of the siile BC, let the two straight lines BD, CD, be drawn to the point D within the triangle Sequence.--1. BD, DC, shall be less than the sides BA, AC. 2. But BD, DC, shall contain an angle, BDC, greater than the angle BAC. Construction.-Produce BD to E. Demonstration.-(I.) 1. Because two sides of a triangle are greater than the third side (I. 20), the two sides BA, AE, of the triangle BAE are greater than BE. 2. To each of these add EC. 3. Therefore the sides BA, AC, are B greater than BE, EC. (ax. 4.) 4. Again, because the two sides CE, ED, of the triangle CED, are greater than CD. (I. 20.) 5. To each of these aud DB. 6. Therefore CE, EB, are greater than CD, DB. (ax. 4.) 7. But it has been shown that BA, AC, are greater than BE, EC. (dem. 3.) 8. Much more then are BA, AC, greater than BD, DC. Demonstration.—(II.) 1. Again, because the exterior angle of a triangle is greater than the interior and opposite angle (I. 16), therefore BDC, the exterior angle of the triangle CDE, is greater than CED. 2. For the same reason, CEB, the exterior angle of the triangle ABE, is greater than the angle BAE. 3. And it has been shown that the angle BDC is greater than CEB. 4. Much more then is the angle BDC greater than the angle BAC. Conclusion.--Therefore, if from the ends, &c. Q. E. D. PROPOSITION 22.—PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these lines must be greater than the third. (I. 20.) (References-Prop. I. 3; post. 3; ax. 1; def. 15.) Given.—Let A, B, C, be the three given straight lines, of which any two whatever are greater than the third; namely, A and B greater than C; A and C greater than B; and B and C greater than A. Sought.-It is required to make a triangle of which the sides shall be equal to A, B, and C, each to each. Construction.--1. Take a straight line DE terminated at the point D, but unlimited towards E. 2. Make DF equal to A, FG equal to B, and GH equal to C. (I. 3.) 3. From the centre F, at the distance FD, describe the circle G H DKL. (post. 3.) 4. From the centre G, at the distance GH, describe the circle HLK. (post. 3.) 5. Join KF, KG. Then the triangle KFG shall have its sides equal to the three straight lines A, B, C. Proof.-1. Because the point F is the centre of the circle LKL, FD is equal to FK. "(def. 15.) 2. But FD is equal to A. (const.) 4. Again, because the point G is the centre of the circle HLK, GH is equal to GK. (def. 15.) 5. But GH is equal to C. (const.)" 7. And FG is equal to B. (const.) 8. Therefore the three straight lines KF, FG, GK, are equal to the three A, B, C, each to each. Conclusion.-Therefore the triangle KFG has its three sides KF, FG, GK, equal to the three given straight lines A, B, C. 'Q. E. F. PROPOSITION 23.—PROBLEM. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. (References, Prop. I. 8, 22.) Given.-Let AB be the given straight line, and A the given point in it, and DCE the given rectilineal angle. Sought.-It is required to make an angle at the point A, in the straight line AB, equal to the rectilineal angle DCE. Construction.-1. In CD, CE, take any points D, E, and join DE. 2. On AB construct a triangle AFG, the sides of which shall be equal to the three straight lines CD, DE, EC; namely, ÅF equal to CD, FG to DE, and GA to EC. (I. 22.) Then the angle FAG shall be equal to the angle DCE. Proof.-1. Because DC, CE, are equal to FA, AG, each to each, and the base DE equal to the base FG; (const.) 2. The angle DĆE is equal to the angle FAG. (I. 8.) Conclusion.—Therefore at the given point A in the given straight line AB, the angle FAG has been made equal to the given rectilineal angle DCE. Q. E. F. PROPOSITION 24.-THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, but the angle contained by the two sides of one of them greater than the angle contained by the two sides equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other. (References, Prop. I. 3, 4, 5, 19, 23; ax. 9.) Hypothesis.—Let ABC, DEF, be two triangles which have |