to the angle GDE, the angle DFC will also be equal to the angle DFE (Prop. 4.) But one line is perpendicular to another when the angles on both fides of it are equal (Def. 8.); therefore CF is perpendicular to AB; and it is drawn from the point c, as was to be done. PROP. XIII. THEOREM. The angles which one right line makes with another, on the fame fide of it, are together equal to two right angles. Let the right line AB fall upon the right line CD; then will the angles ABC, ABD, taken together, be equal to two right angles. For if the angles ABC, ABD be equal to each other, they will be, each of them, right angles (Def. 8 and 9.) But if they be unequal, let EB be drawn, from the point в, perpendicular to CD (Prop. 11.) Then, fince the angles EBC, EBD are right angles. (Def. 8.), and the angle EBD is equal to the angles EBA, ABD (Ax. 8.), the angles EBC, EBA and ABD will be equal to two right angles. But the angles EBC, EBA are, together, equal to the angle ABC (Ax. 8.); confequently the angles ABC, ABD are, alfo, equal to two right angles. Q. E. D. COROLL. All the angles which can be made, at any point в, on the fame fide of the right line CD, are, together, equal to two right angles. xiv. PRO P. XIV. THEOREM. If a right line meet two other right lines, in the fame point, and make the angles on each fide of it together equal to two right angles, thofe lines will form one continued right line. Ε B Let the right line AB meet the two right lines CB, BD, at the point B, and make the angles ABC, ABD together equal to two right angles, then will BD be in the fame right line with CB. For, if it be not, let fome other line BE be in the fame. right line with CB. Then, because the right line AB falls upon the right line CBE, the angles ABC, ABE, taken together, are equal to two right angles (Prop. 13.) But the angles ABC, ABD are also equal to two right angles (by Hyp.); confequently the angles ABC, ABE are equal to the angles ABC, ABD. And, if the angle ABC, which is common, be taken away, the remaining angle ABE will be equal to the remaining angle ABD; the less to the greater, which is abfurd. The line BE, therefore, is not in the fame right line with CB; and the fame may be proved of any other line but BD; confequently CBD is one continued right line, as was to be fhewn. PROP. XV. THEOREM. If two right lines interfect each other, the oppofite angles will be equal, E Let the two right lines AB, CD interfect each other in the point E; then will the angle AEC be equal to the angle DEB, and the angle AED to the angle CEB. For, fince the right line CE falls upon the right line AB, the angles AEC, CEB, taken together, are equal to two right angles (Prop. 13.) And, because the right line BE falls upon the right line CD, the angles BED, CEB, taken together, are also equal to two right angles (Prop. 13.) The angles AEC, CEB, taken together, are, therefore, equal to the angles BED, CEB taken together (Ax. 1.) And, if the angle CEB, which is common, be taken away, the remaining angle AEC will be equal to the remaining angle BED (Ax. 3.) And, in the fame manner, it may be fhewn that the angle AED is equal to the angle CEB. Q. E. D. COROLL. All the angles made by any number of right lines, meeting in a point, are together equal to four right angles. PROP. XVI. THEOREM. If one fide of a triangle be produced, the outward angle will be greater than either of the inward opposite angles. D Let ABC be a triangle, having the fide AB produced to D; then will the outward angle CBD be greater than either of the inward oppofite angles BAC or ACB. For, bifect BC in E (Prop. 10.), and join AE; in which, produced, take EF equal to AE (Prop. 3.), and join BF. Then, fince AE is equal to EF, EC to EB (by const.), and the angle AEC to the angle BEF (Prop. 15.), the angle ACE will, alfo, be equal to the angle EBF (Prop. 4.) But the angle CBD is greater than the angle EBF; confequently it is also greater than the angle ACE. And, if CB be produced to G, and AB be bisected, it may be fhewn, in like manner, that the angle ABG, or its equal CBD, is greater than CAB. Q. E. D. PRO P. XVII. THEOREM. The greater fide of every triangle is oppofite to the greater angle; and the greater angle to the greater fide. D B Let ABC be a triangle, having the fide AB greater than the fide AC; then will the angle ACB be greater than the angle ABC. For, fince AB is greater than AC, let AD be taken equal to AC (Prop. 3.), and join CD. Then, fince CDB is a triangle, the outward angle ADC is greater than the inward oppofite angle DBC (Prop. 16.) But the angle ACD is equal to the angle ADC, because AC is equal to AD; confequently the angle ACD is, also, greater than DBC or ABC. And, fince ACD is only a part of ACB, the whole angle ACB must be much greater than the angle ABC. Again, let the angle ACB be greater than the angle ABC, then will the fide AB be greater than the fide AC. |