the equal sides are opposite. Wherefore the angle BAE is equal to the angle ECF. But the angle ECD is greater (Ax. 9) than the angle ECF. Therefore the angle A CD is greater than the angle B A E. In the same manner, if the side B C be bisected, and A C be produced to G, it may be demonstrated that the angle BCG, is greater than the angle AB Č. But the angle ACD is equal (I. 15) to the angle BCG. Therefore the angle ACD is greater than the angle ABC. Therefore, if one side, &c. Q. E. D. The student should for the sake of practice, write out the demonstration of the PROP. XVII. THEOREM. Any two angles of a triangle are together less than two right angles. Let ABC be any triangle; any two of its angles are together less than two right angles. B Produce B C to D. Because ACD is the exterior angle of the triangle A B C, the angle A CD is greater (I. 16) than the interior and opposite angle ABC. To each of these unequals, add the angle ACB. Therefore the two angles ACD, ACB, are greater (Ax. 4) than the two angles ABC, ACB. But the two angles ACD, ACB are together equal (I. 13) to two right angles. Therefore the two angles ABC, BCA are together less than two right angles. In like manner, it may be demonstrated, that the two angles BAC, ACB, as also the two angles CAB, ABC, are together less than two right angles. Therefore any two angles, &c. Q. E. D. Exercise 1.-The three interior angles of any triangle are less than three right angles. Exercise 2.-The two exterior angles of any triangle are greater than two right angles; and the three exterior angles are greater than three right angles. PROP. XVIII. THEOREM. The greater side of every triangle is opposite to the greater angle. Let A B C be a triangle, of which the side AC is greater than the side AB. greater than the angle B C A. From A C cut off (I. 3) AD BD. A The angle ABC is equal to AB. Join D B Because ADB is the exterior angle of the triangle BDC, it is greater (I. 16) than the interior and opposite angle DCB. But the angle ADB is equal (I. 5) to the angle ABD, because the side AB is equal (Const.) to the side AD. Therefore the angle ABD is likewise greater than the angle ACB. Much more, therefore, is the angle ABC greater than the angle ACB. Therefore the greater side, &c. Q. E. D. C Corollary.-One angle of a triangle is greater than, equal to, or less than another, according as the side opposite to the former is greater than, equal to, or less than the side opposite to the latter. Corollary 2.-All the angles of a scalene triangle are unequal. PROP. XIX. THEOREM. The greater angle of every triangle is the greater side subtended by the greater side, or has opposite to it. Let A B C be a triangle, of which the angle A B C is greater than the angle BCA. The side A C is greater than the side A B. A For, if the side AC be not greater than the side A B, A C must either be equal to A B, or less than it. The side A C is not equal to the side A B. Because then the angle ABC would be equal (I. 5) to the angle ACB. But it is (Hyp.) not. Therefore AC is not equal to A B. The side AC is not less than the side AB. Because then the angle ABC would be less (I. 18) than the angle ACB. But it is (Hyp.) not. Therefore the side AC B is not less than À B. And it has been shown that AC is not equal to AB. Therefore the side AC is greater than the side AB. Wherefore the greater angle, &c. Q. E. D. Corollary.-One side of a triangle is greater than, equal to, or less than another, according as the angle opposite to the former, is greater than, equal to, or less than the angle opposite to the latter. Exercise. If from a point without a given straight line, any number of straight lines be drawn to meet it; of all these straight lines, that which is perpendicular to the given straight line is the least; and of others that which is nearer to the perpendicular is always less than the more remote; also, from the same point only two equal straight lines can be drawn to the given straight line, one upon each side of the shortest line. 3 PROP. XX. THEOREM. Any two sides of a triangle are together greater than the third side. Let A B C be a triangle: any two of its sides are together greater than the third side; viz., the sides BA, AC, are greater than the side B C; the sides A B, BC greater than AC; and the sides BC, CA greater than A B. Produce B A to the point D, and make (I. 3) AD equal to AC. Join D C. B D Because DA is equal to AC, the angle ADC is equal (I. 5) to the angle A CD. But the angle BCD is greater (Ax. 9) than the angle ACD: therefore the angle B CD is greater than the angle AD C. But because the angle BCD of the triangle DCB is greater than its angle B D C, and the greater (I. 19) angle is subtended by the greater side. Therefore the side D B is greater than the side B C. Again AD is equal (Const.) to A C. To each of these equals, add B A. Therefore the whole BD is equal (Ax. 2) to the two B A and A C. But BD was proved to be greater than B C. Therefore the sides BA, AC are greater than B C. In the same manner it may be demonstrated, that the two sides AB, BC are greater than CA, and the two sides BC, CA greater than AB. Therefore any two sides, &c. Q. E. D. This proposition is a corollary to the definition of a straight line, given in the annotation to Def. 4. Exercise 1.-Any side of a triangle is greater than the difference between the other two sides. Exercise 2.-The three sides of a triangle taken together are greater than the double of any one side, but less than the double of any two sides. Exercise 3.-From two given points on the same side of a straight line given in position, to draw two straight lines which shall meet at a point in it, and which taken together shall be less than the sum of two straight lines drawn from the same points to any other point in the given straight line. PROP. XXI. THEOREM. If from the ends of one side of a triangle, there be drawn two straight lines to a point within the triangle, these together shall be less than the other two sides of the triangle, but shall contain a greater angle. Let A B C be a triangle, and from the points B, C, the ends of the side B C, let the two straight lines BD, CD be drawn to the point D within the triangle. Then BD and D C together shall be less than the other two sides BA, AC of the triangle AB C, but shall contain an angle BDC_greater than the angle BAC. A D Produce B D to E. Then, the two sides BA, A E, of the triangle ABE are greater than the third side B E (I. 20). To each of these unequals add E C. Therefore the two sides BA, AC are greater (Ax. 4) than BE, EC. Again, the two sides CE, ED of the triangle CED. are greater (I. 20) than the third side CD. To each of these unequals, add D B. Therefore the two sides CE, E B, are greater (Ax. 4) than CD, DB. But it has been shown that B A, AC are greater than BE, EC. B Much more then are B A, AC, greater than BD, DC. Again, the exterior angle BDC of the triangle CDE is greater than its interior and opposite angle CED (I. 16). And the exterior angle CEB of the triangle AB E is greater than its interior and opposite angle BAC (I. 16). But the angle BDC is greater than the angle CEB. Much more then is the angle BDC greater than the angle BAC. Therefore, if from the ends of, &c. Q. E. D. Exercise. If from any point within a triangle, straight lines be drawn to the vertices of the three angles, these three straight lines taken together shall be less than the sum of the three sides, but greater than half that sum. PROP. XXII. PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of them must be greater than the third (I. 20). Let A, B, C, be the given straight lines, of which any two whatever are greater than the third; viz., A and B greater than C; A and C greater than B; and, B and C greater than A. It is required to make a triangle of which the sides shall be equal to A, B, C, each to each. Take a straight line DE terminated at the point D, but unlimited towards E. Make (I. 3) DF equal to A, FG equal to B, and GH equal to C. From the centre F, at the distance FD, describe (Post. 3) the circle DKL. From the centre G, at the distance GH, describe (Post. 3) another circle HLK. And join KF, KG. The triangle KFG has its sides equal to the three straight lines A, B, C. K D LA H Because the point F is the centre of the circle D KL, FD is equal (Def. 15) to FK. But FD is equal (Const.) to the straight line A. Therefore FK is equal (Ax. 1) to A. Again, because G is the centre of the circle L KH, GH is equal (Def. 15) to GK. But G H is equal to C. Therefore also GK is equal to C. And FG is equal (Const.) to B. Therefore the three straight lines KF, FG, GK, are equal to the three straight lines A, B, C. Wherefore the triangle KFG has been made, having its three sides KF, FG, GK, equal to the three given straight lines A, B, C. Q. E. F. This is the general proposition of which Prop. I. is but a particular case. It is evident that upon the other side of the base FG, another triangle might be constructed, having its three sides equal to the three given straight lines. In the demonstration, it is assumed that the two circles will intersect each other. To prove this, it is sufficient to observe that the sum of the radii of the two circles is, by hypothesis, greater than the distance between their centres. Exercise 1.-To make a triangle equal to a given triangle. Exercise 2.—To make a rectilineal figure equal to a given rectilineal figure. PROP. XXIII. PROBLEM. At a given point in a given straight line, to make a rectilineal angle equal to a given rectilineal angle. Let AB be the given straight line, A the given point in it, and DCE the given rectilineal angle. It is required to make an angle at the given point A in the given straight line AB, that shall be equal to the given rectilineal angle DCE. D A G F B In CD, CE, take any points D, E, and join DE. Upon the straight line AB make (I. 22) the triangle AF G, the sides of which shall be equal to the three straight lines CD, DE, EC, that is, AF equal to CD, AG to CE, and FG to DE. The angle FA G is equal to the angle DCE. Because the two sides FA, AG, are equal to the two sides DC, CE, each to cach, and the base FG to the base DE. Therefore the angle FAG is equal (I. 8) to the angle DCE. Wherefore at the given point A, in the given straight line AB, the angle FAG is made equal to the given rectilineal angle DCE. Q. E. F. It is evident that upon the other side of the straight line AB, another angle might be made equal to the given angle DCE. Exercise.-At a given point in a given straight line, to make an angle equal to the supplement of a given angle; also, to make an angle equal to its complement. 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 is greater than the base of the other. Let ABC, DEF be two triangles, which have the two sides AB, AC, equal to the two sides DE, DF, each to each; viz., AB equal to DE, and AC to DF. But the angle BAC greater than the angle EDF. The base BC is greater than the base EF. Of the two sides DE, DF, let DE be the side which is not greater than the other. At the point D, in the straight line DE, make (I. 23) the angle EDG equal to the angle BAC. Make D G equal (1.3) to AC or DF. And join EG, GF. F Because D E is equal (Hyp.) to AB, and DG (Const.) to AC, the two sides, ED, DG, are equal to the two BA, A C, each to each. And the angle EDG is equal (Const.) to the angle BAC. Therefore the base EG is equal (I. 4) to the base BC. Again, because DG is equal to DF, the angle DFG is equal (I. 5) to the angle DGF. But the angle DGF is greater (Ax. 9) than the angle EGF. Therefore the angle DFG is also greater than EGF. Much more then is the angle EFG greater than the angle EGF. Now, because the angle EFG of the triangle EFG is greater than its angle EGF, and the greater (I. 19) angle is subtended by the greater side. Therefore the side EG is greater than the side EF. But EG was proved to be equal to BC. Therefore BC is greater than EF. Therefore if two triangles, &c. Q. E. D. Dr. Simson in the construction of this proposition, introduced these words: "of the two sides DE, DF, let D E be the side which is not greater than the other," in order to avoid three distinct cases of construction, which would arise by taking that side which is greater than the other. Exercise 1.-Demonstrate this proposition, by making the construction on the greater of the two sides of the triangle DEF, and exhibit the three distinct cases above mentioned. Exercise 2.-Demonstrate that in Dr. Simson's construction, the straight line EG cuts the straight line D F in some point between D and F. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other; the angle contained by the two sides of that which has the greater base, is greater than the angle contained by the two sides equal to them of the other. Let ABC, DEF be two triangles which have the two sides AB, AC equal to the two sides DE, DF, each to each; viz., AB equal to DE |