2. The equilateral triangle may be described on either side of these two lines, giving four equilateral triangles. 3. In each of these possible triangles the sides may be produced either through the vertex or through the base, thus giving eight variations of the figure. Ex. 1. 2. Construct and demonstrate this problem by joining A with Draw the figure of this proposition when the given point PROPOSITION 3. PROBLEM. From the greater of two given straight lines, to cut off a part equal to the less. Let AB and C be the two given straight lines, of which AB is the greater. It is required to cut off from AB, the greater, a part equal to Ɑ the less. From the point 2. A draw the straight line AD equal to C; [I. 2.] and from the centre A, at the distance AD, describe the circle DEF. [Post. 3.] AE shall be equal to C. B Because the point A is the centre of the circle 3. DEF, AE is equal to AD. [Def. 15.] But C is equal to AD. [Const.] Therefore AE and Care each of them equal to AD. Therefore AE is equal to C. [Ax. 1.] Therefore from AB the greater of two given straight lines, a part AE has been cut off equal to C the less. Q.E.F. If Euclid had permitted the use of compasses, which would transfer distances, this problem would have been solved at once by marking off the smaller line on the larger. As it is, its solution requires the description of five circles. The reason of this is also explained on p. 8. Ex. Show how to produce the less of two lines until it is equal to the greater. PROPOSITION 4. THEOREM. If two triangles have two sides of the one equal to two sides of the other, each to each, and have also the angles contained by those sides equal to one another, they shall also have their bases, or third sides equal; and the two triangles shall be equal, and their other angles shall be equal, each to each, namely, those to which the equal sides are opposite. A Let ABC, DEF be two triangles which have the 1. two sides AB, AC equal to the two sides DE, DF, each to each, namely, AB to DE and AC to DF; and the angle BAC equal to the angle EDF, the base BC shall be equal to the base EF, and the triangle ABC to the triangle DEF, B and the other angles shall be equal each to each, to which the equal sides are opposite, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. For if the triangle ABC be applied to the 2, 3. triangle DEF, so that the point A may be on the point D, and the straight line AB on the straight line DE, the point B will coincide with the point E, because AB is equal to DE. [Hyp.] And, AB coinciding with DE, AC will coincide with DF, because the angle BAC is equal to the angle EDF. [Hyp.] Therefore also the point will coincide with the point F, because AC is equal to DF: [Hyp.] But the point B was proved to coincide with the point E, therefore the base BC will coincide with the base EF; because, B coinciding with E and C with F, if the base BC does not coincide with the base EF, two straight lines will enclose a space; which is impossible. [Ax. 10.] Therefore the base BC coincides with the base EF, and is equal to it. [Ax. 8.] Therefore the whole triangle ABC coincides with the whole triangle DEF, and is equal to it. [Ax. 8] And the other angles of the one coincide with the other angles of the other, and are equal to them, namely, the angle ABC to the angle DEF, and the angle ACB to the angle DFE. Therefore, if two triangles, &c. Q.E.D. The letters Q.E.D.= quod erat demonstrandum, which was to be proved, they are usually added at the close of a theorem. The method of proof here employed is called proof by 'superposition.' Note how carefully Euclid fits one triangle upon the other. First he shows that AB exactly coincides with DE. Next, that the angle BAC coincides with EDF. Thirdly, that the line AC coincides with DF. And lastly, from these he shows that BC must coincide with EF. Hence the triangle ABC coincides with DEF. This is one of the most important results in Book I. Ex. 1. Prove by the method used in this proposition that the two triangles into which a square is divided by its diagonal are equal to one another. 2. The diagonals of a square are equal to one another. PROPOSITION 5. THEOREM. The angles at the base of an isosceles triangle are equal to one another; and if the equal sides be produced the angles on the other side of the base shall be equal to one another. Let ABC be an isosceles triangle, having the 1. side AB equal to the side AC, and let the straight lines AB, AC be produced to D and E; the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE. 2. In BD take any Because A Fis equal And because the whole AF is equal to the whole AG, of which the parts AB, AC are equal, [Hyp.] the remainder BF is equal to the remainder CG. [Ax. 3.] And FC was proved to be equal to GB; therefore the two sides BF, FC are equal to the two sides CG, GB, each to each; and the angle BFC was proved equal to the angle CGB; therefore the triangles BFC, CGB are equal, and their other angles are equal, each to each, to which the equal sides are opposite, namely the angle FBC to the angle GCB, and the angle BCF to the angle CBG. [I. 4.] And since it has been proved that the whole angle ABG is equal to the whole angle ACF, and that the parts of these, the angles CBG, BCF are also equal; therefore the remaining angle ABC is equal to the remaining angle ACB, which are the angles at the base of the triangle ABC. [Ax. 3.] And it has also been proved that the angle FBC is equal to the angle GCB, which are the angles on the other side of the base. Therefore the angles, &c. Q.E.D. Corollary. Hence every equilateral triangle is also equiangular. If the pupil finds serious difficulty in mastering this proposition, it will probably be a help to distinguish the triangles spoken of by shading. Or two figures may be drawn, one to show the triangle AFC, and another to show the triangle ABG. The same method may be employed with BFC, BGC. But as a final exercise the proposition should be mastered as it stands. A Corollary to a theorem is a truth easily deduced from that theorem. Ex. 1. Take F in AB, and making AG equal to AF, show that PROPOSITION 6. THEOREM. If two angles of a triangle be equal to one another, the sides also which subtend, or are opposite to, the equal angles, shall be equal to one another. 1. Let ABC be a triangle, having the angle ABC equal to the angle ACB. The side AC shall be equal to the side AB. A 2. For if AC be not equal to AB, one of them is greater than the other. Let AB be the greater, and from it cut off DB equal to AC the less, [I. 3.] and join DC. B 3. Then, because in the triangles DBC, ACB, DB is equal to AC, [Con.] and BC is common to both, the two sides DB, BC are equal to the two sides |