and, taking away the common angle ACD, we have, ACE DCB. Hence, the proposition is proved. Cor. 1. If one of the angles about C all of the others will be right angles also. is a right angle, For, (P. I., C. 1), D A E Cor. 2. If one line DE, is perpendicular to another AB, then will the second line AB be perpendicular to the first DE. and DCB are right angles, by For, the angles DCA definition (D. 12); and from what has just been proved, the angles ACE and BCE are also right angles. Hence, the two lines are mutually perpendicular to each other. Cor. 3. The sum of all the angles ACB, BCD, DCE, ECF, FCA, that can be formed about a point, is equal to four right angles. -D * F For, if two lines be drawn through the point, mutually perpendicular to each other, the sum of the angles which they form will be equal to four right angles, and it will also be equal to the sum of the given angles (A. 9). Hence, the sum of the given angles is equal to four right angles. PROPOSITION III. THEOREM. If two straight lines have two points in common, they will coincide throughout their whole extent, and form one and the same line. coincide (A. 11). Suppose, now, that they begin to separate at some point C, beyond AB, the one becoming ACE, and the other ACD. If the lines do separate at C, one or the other must change direction at this point; but this is contradictory to the definition of a straight line (D. 4): hence, the supposition that they separate at any point is absurd. They must, therefore, coincide throughout; which was to be proved. Cor. Two straight lines can intersect in only one point. NOTE. The method of demonstration employed above, called the reductio ad absurdum. It consists in assuming an hypothesis which is the contradictory of the proposition to be proved, and then continuing the reasoning until the assumed hypothesis is shown to be false. Its contradictory is thus proved to be true. This method of demonstration is often used in Geometry. PROPOSITION IV. THEOREM. If a straight line meet two other straight lines at a com mon point, making the sum of the contiguous angles equal to two right angles, the two lines met will form one and the same straight line. Let DC meet AC and BC. at C, making the sum of the angles DCA and DCB equal to two right angles: then will CB be the prolongation of AC. D A -B E For, if not, suppose CE to be the prolongation of AC; then will the sum of the angles equal to two right angles (P. I.): have (A. 1), DCA and DCE be DCA+ DCB = DCA + DCE; Taking from both the common angle DCA, there remains, DCB DCE, which is impossible, since a part cannot be equal to the whole (A. 8). Hence, CB must be the prolongation of AC; which was to be proved. PROPOSITION V. THEOREM. If two triangles have two sides and the included angle of the one equal to two sides and the included angle of the other, each to each, the triangles will be equal in all their parts. In the triangles ABC and DEF, let AB be equal to DE, AC to DF, and the angle A to the angle D: then will the triangles be equal in all their parts. For, let ABC be applied to DEF, in such a manner that the angle A shall coincide with the angle D, the side AB taking the direction DE, and A A B the side AC the direction DF. Then, because AB is equal to DE, the vertex B will coincide with the vertex E; and because AC is equal to DF, the vertex C will coincide with the vertex F; consequently, the side BC will coincide with the side EF (A. 11). The two triangles, therefore, coincide throughout, and are consequently equal in all their parts (I., D. 14); which was to be proved. PROPOSITION VI. THEOREM. If two triangles have two angles and the included side of the one equal to two angles and the included side of the other, each to each, the triangles will be equal in all their parts. will the triangles be equal in all their parts. For, let ABC be applied to DEF in such a manner that the angle B shall coincide with the angle E, the side BU taking the direction EF, and the side BA the direc tion ED. Then, because BC is equal to EF, the vertex C will coincide with the vertex F; and because the angle C is equal to the angle F, the side CA will take the direction FD. Now, the vertex A being at the same time on the lines ED and FD, it must be at their intersection D (P. III., C.): hence, the triangles coincide throughout, and are therefore equal in all their parts (I., D. 14); which was to be proved. PROPOSITION VII. THEOREM. The sum of any two sides of a triangle is greater than the third side. Let ABC be a triangle: then will the sum of any two sides, as AB, BC, be greater than the third side AC. For, the distance from A to C, measured on any broken line AB, BC, is greater than the distance measured on the straight line AC (A. 12): hence, the sum of AB and BC is greater than AC; which was to be proved. Cor. If from both members of the inequality, ACAB+ BC, we take away either of the sides AB, BC, example, there will remain (A. 5), as BC, for that is, the difference between any two sides of a triangle is less than the third side. Scholium. In order that any three given lines may re |