Take away the common angle AED, and (Ax. 3.) 4. The remaining angle CEA is equal to the remaining angle DEB. In the same manner it can be demonstrated that 5. The angles CEB, 4ED, are equal. Therefore, if two straight lines, &c. Q.E.D. COR. 1.-From this it is manifest, that, if two straight lines cut one another, the angles they make at the point where they cut, are together equal to four right angles. COR. 2.-And consequently that all the angles made by any number of lines meeting in one point are together equal to four right angles. PROP. XVI.-THEOREM. If one side of a triangle be produced, the exterior angle is greater than either of the interior opposite angles. Let ABC be a triangle, and let its side BC be produced to D, the exterior angle ACD is greater than either of the interior opposite angles CBA, BAČ. B F E Bisect (I. 10.) AC in E, join BE and produce it to F, and make EF equal to BE; join also FC, and produce AC to G. Because AE is equal to EC, and BE to EF; 1. and (I. 15.) 2. AE, EB, are equal to CE, EF, each to each; The angle AEB is equal to the angle CEF, because they are opposite vertical angles; therefore (I. 4.) 3. The base AB is equal to the base CF, and the triangle AEB to the triangle CEF, and the remaining angles to the remaining angles, each to each, to which the equal sides are opposite: wherefore, 4. The angle BAE is equal to the angle ECF; but the angle ECD is greater than the angle ECF; therefore In the same manner, if the side BC be bisected, it may be demonstrated that the angle BCG that is (I. 15.) 6. The angle ACD is greater than the angle ABC. Therefore if one side, &c. q.E.D. 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 together are less than two right angles. A B Produce BC to D; and because ACD is the exterior angle of the triangle ABC, (I. 16.) 1. ACD is greater than the interior and opposite angle ABC; to each of these add the angle ACB; therefore 2. The angles ACD, ACB, are greater than the angles ABC, ACB; ACD, ACB, are together equal to two right angles; 4. The angles ABC, BCA, are less than two right angles. In like manner, it may be demonstrated, that 5. BAC, ACB, as also CAB, ABC, are less than two right angles. Therefore any two angles, &c. Q.E.D. The greater side of every triangle is opposite to the greater angle. Let ABC be a triangle, of which the side AC is greater than the side AB; the angle ABC' is also greater than the angle BCA. B Because AC is greater than AB, make (I. 3.) AD equal to AB, and join BD; and because ADB is the exterior angle of the triangle BDC, (I. 16.) 1. ADB is greater than the interior and opposite angle ᎠᏟᏴ ; but (I. 5.) 2. ADB is equal to ABD, because the side AB is equal to the side AD; therefore likewise 3. The angle ABD is greater than the angle ACB, wherefore much more 4. The angle ABC is greater than ACB. Therefore the greater side, &c. Q.E.D. PROP. XIX.-THEOREM. The greater angle of every triangle is subtended by the greater side, or has the greater side opposite to it. Let ABC be a triangle, of which the angle ABC is greater than the angle BCA; the side AC is likewise greater than the side AB. B For if it be not greater, AC must either be equal to AB, or less than it; it is not equal, because then the angle ABC would be equal (I. 5.) to the angle ACB; but it is not; therefore 1. AC is not equal to AB; neither is it less; because then the angle ABC would be less (I. 18.) than the angle ACB; but it is not; therefore 2. The side AC is not less than AB ; and it has been shown that it is not equal to AB; therefore 3. AC is greater than AB. Wherefore the greater angle, &c. Q.E.D. PROP. XX.-THEOREM. Any two sides of a triangle are together greater than the third side. Let ABC be a triangle; any two sides of it together are greater than the third side, viz. the sides BA, AC, greater than the side BC; and AB, BC, greater than ɅC; and BC, CA, greater than AB. B Produce BA to the point D, and make (I. 3.) AD equal to AC; and join DC. Because DA is equal to AC, likewise (I. 5.) 1. The angle ADC is equal to ACD; but the angle BCD is greater than the angle ACD; therefore 2. The angle BCD is greater than the angle ADC; and because the angle BCD of the triangle DCB is greater than its angle BDC, and that the greater (I. 19.) side is opposite to the greater angle; therefore 3. The side DB is greater than the side BC; but DB is equal to B▲ and AC; therefore 4. The sides BA, AC, are greater than BC. In the same manner it may be demonstrated that 5. The sides AB, BC, are greater than CA; and BC, CA, are greater than AB. Therefore any two sides, &c. Q.E.D. PROP. XXI.-THEOREM. If from the ends of the side of a triangle, there be drawn two straight lines to a point within the triangle, these shall be less than the other two sides of the triangle, but shall contain a greater angle. Let the two straight lines BD, CD, be drawn from B, C, the ends of the side BC of the triangle ABC, to the point D within it; BD and DC are less than the other two sides BA, AC, of the triangle, but contain an angle BDC greater than the angle BAC. A E D B Produce BD to E; and because two sides of a triangle are greater than the third side (I. 20.) 1. The two sides BA, AE, of the triangle ABE, are greater than BE; to each of these add EC; therefore 2. The sides BA, AC, are greater than BE, EC. Again, because (I. 20.) 3. than CD, The two sides CE, ED, of the triangle CED, are greater add DB to each of these; therefore (Ax. 4.) 4. The sides CE, EB, are greater than CD, DB; but it has been shown that BA, AC, are greater than BE, EC; much more then 5. BA, AC, are greater than BD, DC. Again, because the exterior angle of a triangle is greater than the interior and opposite angle (I. 16.) 1. than CED. The exterior angle BDC of the triangle CDE is greater For the same reason, 2. The exterior angle CEB of the triangle ABE is greater than BAC; and it has been demonstrated that the angle BDC is greater than CEB; much more then 3. The angle BDC is greater than the angle BAC. Therefore, if from the ends of, &c. Q.E.D. PROP. XXII.-PROBLEM. To make a triangle of which the sides shall be equal to three given straight lines, but any two whatever of these must be greater than the third, (I. 20.) Let A, B, C, be the three given straight lines, of which any two whatever are greater than the third; viz. A and B greater than C; 4 and C greater than B; and B and C than 4. 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, and make (I. 3.) DF equal to A, FG equal to B, and GH equal to C; and from the centre F, at a distance FD, describe (Post. 3.) the circle DKL; and from the centre G, at the distance GH, describe another circle HLK; and join KF, KG; the triangle KFG has its sides equal to the three straight lines A, B, C. Because the point F is the centre of the circle DKL, (Def. 15.) 1. FD is equal to FK; but FD is equal to the straight line 4; therefore |