The method usually followed in the propositions is 1. A general statement or enunciation of the theorem to be proved or problem to be solved. 2. A restatement of it as applied to the particular figure on the paper. 3. The construction of any additional lines which may be needed to prove the proposition. 4. The proof, which terminates with the statement which we wished to prove. 5. Any corollaries or scholiums which may attach to the proposition. Proposition 1. Problem. From the greater of two given straight lines, to cut off a part equal to the less. Let AB and C be two straight lines, AB being the Restatement. greater: it is required to cut off from AB a part equal to C. With the centre A, and radius equal to C, describe (Post. 3) a circle, cutting AB in D. Construction. Proof. NOTE.—A small arc cutting AB in D is sufficient in practice. Restatement. than the third. Proof. Proposition 2. Theorem.-Two sides of a triangle are together greater than the third side. / Let ABC be a triangle; any two sides are together greater D AD is a radius of the circle, and is therefore equal (Def. 30) to C. Because (Ax. 12) the shortest distance between two points is B B C the straight line which joins them, the line BC is the shortest distance from B to C, and is therefore less than BA and AC. Therefore BA and AC are together greater than BC. In the same way, any other two sides may be shown to be together greater than the third. Proposition 3. Problem.-To construct a triangle, having given the three sides. Restatement. It is a necessary condition that any two of the sides be greater than the third. Construction. Let A, B, C be the three sides of a triangle: it is required to construct it. Lay down the indefinite line DE, and from it cut off (I. 1) DF equal to A. With Das a centre, and radius equal to B, describe (Post. 3) the circle GHK. With Fas a centre, and radius equal to C, describe (Post. 3) the circle GKL. From G, where these circles cut, draw (Post. Proof. 1) the lines GD, GF. The triangle GDF is the required triangle. Because DG is a radius of the circle GHK, it is equal (Def. 30) to B; and because FG is a radius of GKL, it is equal to C; and DF was made equal to A; therefore GDF is a triangle having its three sides respectively equal to A, B, C. Scholium. In practice we need not describe the whole circle. Two small arcs intersecting at G will be sufficient. Corollary.-If the triangle is to be equilateral (Def. 16), it is only necessary to know one side, AB. From A and B, with radii equal to AB, describe (Post. 3) arcs of circles cutting in C. ABC is an equilateral triangle. Proposition 4. Theorem.-If two triangles have two sides and the included angle of one, equal to two sides and the included angle of the other, each to each, the triangles will be equal in all their parts. Let ABC, DEF be two triangles which have BA, AC, and the angle A, equal to ED, DF' and the angle D, each to each; then will the triangles be equal in all their parts. + A B Let the triangle ABC be applied* to the triangle DEF, so that A shall be on D, and AB on DE. Then, because AB is equal to DE, B will coincide with E; also because the angle BAC is equal to the angle EDF, AC will coincide with DF, B and because AC is equal to DF, C will coincide with F. Hence, B coinciding with E, and C with F, BC will coincide with EF, and the triangles will coincide, and are therefore equal (Ax. 8) in all their parts. CE D F Scholium.―This proposition shows that if we have two triangles of any shape, in which we can find two sides of one equal to two sides of the other, and also the angles contained by these equal sides equal to each other, it is safe to conclude that the triangles are every way equal. It is important to remember that those angles will be equal which are opposite to the equal sides. Thus, suppose we know that in the two triangles ABC and ABD, AB is the same in both, AC is equal to AD, and the angle CAB, included by the sides CA and AB, is equal to the angle DAB, included by the sides DA and AB. We thence infer, from the preceding proof, that the triangles are equal in the following respects: 1. Their areas are equal. A 2. Their other sides, CB and DB, are equal to each other. 3. Their other angles are equal to each other, viz., the angles ACB and ADB, which are opposite the common side AB, are equal to each other; and the angles ABC and ABD, which are opposite the equal sides AC and AD, are equal to each other. B D Proposition 5. Theorem.-If two triangles have two angles and the included side of one, equal to two angles and the included side of the other, each to each, the triangles will be equal in all their parts. Let ABC, DEF be two triangles which have the two angles ABC, ACB, and the side BC, respectively equal to DEF, DFE, and the side EF. The triangles will be equal in all their parts. B D AA CE Apply the triangle ABC to the triangle DEF, so that B will be on E, and BC on EF; and because BC is equal to EF, C will coincide with F. Because the angle ABC is equal to the angle DEF, BA will coincide with ED; and because the angle BCA is equal to the angle EFD, CA will coincide with FD; hence the point A, which is the intersection of the two lines BA and AC, will agree with the point D, which is F the intersection of the two lines ED and DF. Therefore the triangles will coincide altogether, and will be equal in all their parts; and the equal sides will be opposite to the equal angles. Proposition 6. Theorem.-If two sides of a triangle are equal to each other, the angles which are opposite to them will also be equal to each other. Let ABC be a triangle having the side AB equal to the side AC; then will the angle ACB be equal to the angle ABC. For if ACB be not equal to ABC, one of them must be the greater. Let ABC be the greater, and suppose BD to be drawn making the angle ABD equal to the angle ACB. B In the two triangles ACB and ABD, AC is equal (Hyp.) to AB, the angle A is common to the two triangles, and, by the supposition, the angle ACB is equal to the angle ABD. Hence the two triangles ABD and ACB have two angles and the included side of one, equal to the two angles and the included side of the other. They are therefore equal (I. 5) in all their parts. Hence the area of ADB is equal to the area of ABC, a part to the whole, which is impossible (Ax. 9). The supposition that one of the angles ABC and ACB is greater than the other thus leads to an impossibility; it is therefore untrue, and ABC is equal to ACB. D NOTE. This method of proof is called an indirect demonstrationsometimes reductio ad absurdum. The theorem is proven to be true by showing that any contrary supposition leads to an absurdity or impossibility. Corollary. Every equilateral triangle is also equiangular. For, all the sides being equal, the three opposite angles must all be equal. |