1 For if they be not equal, one of them must be greater than the other ; let Efd be the greater; and make the angle EFB equal to AEF (Prop. 20.) Then, fince AB, CD are parallel, the right line FB, which intersects cd, being produced, will meet AB in some point B (Pof. 4.) And, fince EFB is a triangle, the outward angle AEF will be greater than the inward opposite angle EFB (Prop. 16.) But the angles AEF, EFB are equal to each other (by Const.) whence they are equal and unequal, at the same time, which is absurd. The angle EFD, therefore, is not greater than the angle AEF; and, in the same manner, it may be thewn that it is not less ; consequently they must be equal to each other. C. E. D. COROLL. Right lines which are perpendicular to one of two parallel right lines, are also perpendicular to the other. PRO P. XXV. THEOREM. If a right line interfect two parallel right lines, the outward angle will be equal to the inward opposite one, on the same side; and the two inward angles, on the fame fide, will be equal to two right angles. Let the right line er intersect the two parallel right lines AB, CD; then will the outward angle eGo be equal to the inward opposite angle GHD; and the two inward angles BGH, GHD will be equal to two right angles. For, fince the right line er intersects the two parallel right lines AB, CD, the angle Agh will be equal to the alternate angle GHD (Prop. 24.) But the angle agh is equal to the opposite angle EGB (Prop. 15.); therefore the angle EGB will, also, be equal to the angle GhD. Again, since the right line eg falls upon the right line EF, the angles EGB, BGH, taken together, are equal. to two right angles (Prop. 13.) But the angle EGB has been shewn to be equal to the angle GHD; therefore, the angles BGH, GHD, taken together, will, also, be equal to two right angles. Q. E. D. D COROLL. COROLL. If a right line intersect two other right lines, and make the two inward angles, on the same fide, together less than two right angles, those lines, being produced, will meet each other. PRO P. XXVI. THEOREM. Right lines which are parallel to the same right line, are parallel to each other. Let the right lines AB, cd be each of them parallel to EF, then will as be parallel to cD. For, draw any right line Gk, cutting the lines AB, EF, CD, in the points G, H and K. Then, because AB is parallel to EF (by Hyp.), and GH interfects them, the angle agh is equal to the alternate angle GHF (Prop. 24.) And because cd is parallel to EF (by Hyp.), and HK interfe&is them, the outward angle ghy is equal to the inward angle HKD (Prop. 25.) But the angle AGH has been shewn to be equal to the angle GhF; therefore the angle AGK is also equal to the angle GKD. And, since the right line gk intersects the two right lines AB, CD, and makes the angle AGK equal to the alternate angle GKD, AB will be parallei to CD, as was to be shewn. PRO P. XXVII. PROBLEM. Through a given point, to draw a right line parallel to a given right line. Let AB be the given right line, and c the given point; it is required to draw a right line through the point c that shall be parallel to AB. Take any point p in AB, and make de equal to DC (Prop. 3.); and from the points C, E, with the distances CD, ED, describe the arcs rs, nm. Then, fince any two sides of the triangle ecD are, together, greater than the third side (Prop. 18.), those arcs will intersect each other (Prop. 19.) Let them interfect at F; and through the points F, c, draw the line GH, and it will be parallel to AB, as was required. For, fince the sides CF, Fe of the triangle Erc are each equal to the side CD, or De, of the triangle CDE, (by Conf.) and Ec is common, the angle Ech will be equal to the angle ced (Prop. 7.) But these are alternate angles; therefore Gh is parallel to AB (Prop. 24.); and it is drawn through the point c, as was to be done. PRO P. XXVIII. THEOREM. If one side of a triangle be produced, the outward angle will be equal to the two inward opposite angles, taken together ; and the three angles of every triangle, taken together, are equal to two right angles. Let ABC be a triangle, having one of its fides AB produced to D; then will the outward angle CBD be equal to the two inward opposite angles BCA, CAB, taken together; and the three angles BCA, CAB and ABC, taken together, are equal to two right angles. For through the point B, draw the right line be parallel to AC (Prop. 28.) Then, because ee is parallel to Ac, and co intersects them, the angle cbe will be equal to the alternate angle BCA (Prop. 24.) And because be is- parallel to AC, and an intersects them, the outward angle EBD will be equal to the inward angle CAB (Prop. 25.) But the angles CBE, EBD are equal to the whole angle ced; therefore the outward angle CBD is equal to the two inward opposite angles BCA, CAB taken together. |