PROPOSITION I. PROBLEM.-TO construct an equilateral triangle upon a given finite straight line (AB). SOLUTION. From the center A, at the distance AB, describe the circle BCD (a), and from the center B, at the distance BA, describe the circle ACE; and from the point C, in which the circles cut one another, draw the straight lines CA, CB to the points A, B (b); then ABC will be the triangle required. (a) Post. 3. DEMONSTRATION. It is evident that the triangle ABC is constructed on the line AB. And it is also equilateral: for, the lines AC and AB being both radii of the same circle, BCD are equal (c), and the lines AB and CB being both radii of the same circle, ACE are equal; then, because the lines AC and CB are both equal to the same line AB, therefore they are equal to each other (d); that is, the three sides AB, AC, and CB are equal, and the triangle ABC is therefore equilateral. SCHOLIUM. In this Prop. the following axiom is tacitly assumed by Euclid, viz. :-"That a circle whose center is in the circumference of another circle, must be partly within that circle, and partly without it, and therefore, that those circles must necessarily cut or intersect each other." PROPOSITION II. PROBLEM. From a given point (A) to draw a straight line equal to a given finite straight line (BC). SOLUTION. From the given point A draw a straight line to either extremity B of the given line (a). Upon AB construct an equilateral triangle (b). From the center B, at the distance BC, describe the circle CGH (c); and produce the straight line DB until it meets the circumference in G (d). From the center D, at the distance DG, describe the circle GKL (c); and produce the straight line DA until it meets the circumference in L (d). The straight line AL is equal to the given line BC. DEMONSTRATION. For the lines DG and DL being both radii of the same circle, GKL are equal (e), and if the equals DB E. (a) Post. 1. Def. 13 and 16. and DA (f) be taken from each respectively, the remainders BG and AL are equal (g); but the lines BG and BC being both radii of the same circle CGH, are equal (e): therefore the lines BC and AL, being both equal to the same line BG, are equal to each other (h). Therefore, from a given point A, a straight line has been drawn equal to a given straight line BC. SCHOLIA. 1. The construction of this problem will somewhat vary according to the relative positions of the point A and the line BC. (e) Def. 13 and 16. (g) Ax. 3. 2. In practice this problem will be solved by measuring the length of the given line BC with a pair of compasses; and then, applying one leg of the compasses to the point A, the other leg will mark the length of the line required to be drawn from A. In geometry, however, such use of the compasses is not permitted. The only way in which they may be employed is that allowed in the third postulate, viz. to describe a circle whose circumference shall pass through a given point about some other given point as a center. The compasses must be supposed to close of themselves whenever removed from the paper, so that no distance can be carried by means of them. This restricted use of the compasses being borne in mind will enable the student to see the necessity of the first three problems in this book. PROPOSITION III. PROBLEM. From the greater of two given straight lines (AB and C) to cut off a part equal to the less. SOLUTION. From either extremity A of the greater given line draw a straight line AD equal to the lesser given line C (a). From the center A, at the distance AD, describe the circle DEF (b), which shall cut off AE equal to the lesser line C. DEMONSTRATION. For the lines AD and AE being both radii of the same circle DEF are equal (c). But AD and C are equal (d); therefore, because AE and C are both equal to the same line AD, they are equal to each other (e); and from AB the greater of two given lines, a part AE has been cut off, equal to С the less. (a) I. 2. Post. 3. Def. 13 and 16. d) Const. Ax. 1. SCHOLIUM. By a similar operation the lesser line could be extended to equal the greater; thus from either extremity of the lesser line let a line be drawn equal to the greater, then about this same extremity as a center describe a circle with a radius equal to the greater line; extend the lesser line to meet the circumference of this circle, and it will equal the greater line. PROPOSITION IV. THEOREM. If two triangles (ABC and DEF) have two sides of the one respectively equal to two sides of the other (DE and DF to AB and AC), and the angles formed by those sides also equal to one another (D to A); [1] their bases or third sides (EF and BC) will be equal; [2] and the angles at the bases, which are opposite to the equal sides, will be equal (E to B and F to C); [3] and the triangles themselves will be equal. 13 E DEMONSTRATION. For, if the triangle ABC be applied to DEF, so that the point A may be on the point D, the point B on the straight line DE, and that AC and DF may lie on the same side; then AB must lie wholly on DE, for otherwise two straight lines would enclose a space (a); and because AB is equal to DE, the point B must coincide with the point E (6). Further, because the angles A and D are equal, the side AC must fall upon the side DF; and because AC is equal to DF, the point C must coincide with the point F. (a) Ax. 10. [1] Therefore, as the points B and C coincide with the points E and F, the base BC must coincide with the base EF, and be equal to it (c); for otherwise two straight lines would enclose a space (a). [2.] And as the sides which form the angles B and C coincide with the sides which form the angles E and F, those angles themselves must coincide, and therefore must be equal (c). [3.] And as the straight lines which contain the triangle ABC coincide with and are equal to the straight lines which contain the triangle DEF, therefore the triangles themselves must coincide, and must therefore be equal (c). SCHOLIA. 1. In the above demonstration (at b) an axiom is assumed the converse of the eighth axiom, namely, "Magnitudes which are equal coincide with one another when similarly placed." 2. In every triangle there are six quantities or magnitudes, namely, the three sides and the three angles; and (except in two particular cases) when any three of these are given, the other three can be found and the triangle determined. If, therefore, two triangles are found to agree in any three of those quantities by which the triangles are determined, it is evident that those triangles must be equal. The following are the only six cases which can occur: 1. The three angles. 2. The three sides. 3. Two sides and the angle between them. 4. Two sides and the angle opposite to one of them. 5. Two angles and the side between them. 6. Two angles and the side opposite to one of them. The first case is one of the two in which the triangle is not determined; for a triangle may have its sides increased or diminished to any extent without altering the magnitude of its angles. The second case is demonstrated in the eighth proposition. The third case is the subject of the present proposition. B The fourth case is the other one in which the triangle is not determined; for it is quite possible to have two triangles having two sides of the one equal to two sides of the other, and one of the opposite angles of the one equal to the similar angle of the other, and yet for the triangles themselves not to be equal. Thus, let ABC be a triangle in which neither A nor C are right angles and A is less than C; then from B as a center, and the distance BC as radius, describe a circle cutting AC in D, and draw DB. Now it is evident that, in the triangles ABC and ABD, we have the two sides AB and BC equal to the two AB and BD, and the opposite angle A the same in both; and yet the two triangles are not equal. The fifth and sixth cases are demonstrated in the twenty-sixth proposition. 3. The application of one figure to another, so as to prove or disprove their coincidence, as made use of in this proposition, is termed superposition. It has been objected to by some mathematicians as not being strictly geometrical; but, as Mr. De Morgan has observed, it requires only the admission of the following postulate, "That any figure may be removed from place to place without alteration of form, and a plane figure may be turned over on the plane." The latter portion, printed in italics, is required for the proof of the fifth proposition. 4. The enunciation of this proposition really includes three distinct propositions, which have been distinguished by separate numbers. The demonstration of the first of these is an example of the negative or indirect proof termed "Reductio ad Absurdum," which consists in proving a proposition by showing that if it is denied an obvious absurdity follows. Concerning this method of proof, see the Introduction. 5. The base of a triangle is the third side of a triangle as distinguished from the other two, without any regard to whether the triangle stands upon it or not. PROPOSITION V. THEOREM. [1] If a triangle (ABC) be isosceles, the angles at the base (ABC and ACB) are equal to one another; [2] and if the equal sides be produced, the angles formed by the produced sides and the base below the same (CBD and BCE) shall be equal. CONSTRUCTION. Produce the equal sides AB and AC (a), and in the produced part of one of them AB take any point F, and from the other cut off AG equal to AF (b). Draw a straight line from Ĉ to F, and from B to G (c). DEMONSTRATION. [2.] In the triangles ACF and ABG, the side AC is equal to the side AB (d), the side AF to the side AG (e), and the angle A is common to both; therefore the base CF is equal to the base BG, the angle ACF to the angle ABG, and the angle F to the angle G (f); then taking the equal lines AC and AB from the equal lines AG and AF, the remainders CG and BF are equal (g); therefore in the triangles BCG and CBF, because the side CG is equal to the side BF, the side BG to the side CF, and the angle G to the angle F, the angle BCG must be equal to the angle CBF (f), which are the angles formed by the produced sides and the base, below the same. [1.] Further in the same triangles the remaining angles must be equal, BCF to CBG (f); and if these equal angles be taken from the equal angles ACF and ABG, the remaining angles, ACB and ABC, will be equal (g), which are the angles at the base of the given triangle. A COROLLARY. Hence every equilateral triangle is also equiangular; for if each side be taken in succession as the base, it may be shown that the angles adjacent to the side so taken are equal. SCHOLIA. 1. This proposition may also be proved in the following manner: if the triangle ABC be turned over on the plane (see scholium 3 to the preceding proposition), so that the position of the point A may be unaltered, while the side AB lies on AC, then, since the angle A is the same in both, the side AC must fall on AB; and because the sides AB and AC are equal (a), the point C will coincide with the point B, the point B with C, the angle ACB with the angle ABC, and the angle BCG with the angle CBF; and therefore [1] the angle ACB will be equal to the angle ABC (b), and [2] the angle BCG equal to the angle CBF (b). 2. The enunciation of this proposition really includes two separate propositions which have been distinguished by numbers. PROPOSITION VI. B G Hypoth. Ax. 8. A THEOREM.-If two angles (B and C) of a triangle (ABC) are equal, the sides (AC and AB) opposite to those angles are also equal. DEMONSTRATION. For if AB be not equal to AC, one of them is greater than the other; let AB be the greater, from it cut off DB equal to AC the less (a), and draw the line DC. Then in the triangles ABC and DBC, because the side AC is equal to the side DB, the base BC common to both, and the angle ACB equal to the angle DBC (6), therefore the tri B (a) I. 3. |