# 4. 1. A B to D and E: the angle ABC shall be equal to the angle ACB, and the angle CBD to the angle BCE. In BD take any point F; and from AE, the greater, cut off AG equal* to AF, the less, and join FC, GB. * 3. 1. Because AF is equal to t AG, and AB to I AC, the two + Constr. sides FA, AC are equal to the two GA, AB, each to each: + Hyp. and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal * to the • 4. 1. base GB, and the triangle AFC to the triangle AGB; and the remaining angles of the one, are equal to the remaining angles of the other, each to each, to which the equal sides are opposite ; viz. the angle ACF to the angle ABG, and the angle AFC to the angle AGB: and because the whole AF is equal to the whole AG, of which the parts AB, AC are equal, the remainder BF is equal to the remainder CG: and FC was proved E . 3 Ax. to be equal to GB; therefore the two sides BF, FC are equal to the two CG, GB, each to each ; and the angle BFC was proved to be equal to the angle CGB, and the base BC is common to the two triangles BFC, CGB; wherefore these triangles are equal, and their remaining angles, each to each, to which • 4. 1. the equal sides are opposite: therefore the angle FBC is equal to the angle GCB, and the angle BCF to the angle CBG. And, since it has been demonstrated, that the whole angle ABG is equal to the whole ACF, the parts of which, the angles CBG, BCF are also equal ; therefore the remaining angle ABC is equal + to the remaining angle ACB, which are the angles at +3 Ax. the base of the triangle ABC: and it has also been proved that the angle FBC is equal to the angle GCB, which are the angles upon the other side of the base. Therefore, the angles at the base, &c. Q. E. D. COROLLARY.-Hence, every equilateral triangle is also equiangular. PROPOSITION VI. THEOR.--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. Let ABC be a triangle, having the angle ABC equal to the angle AGB : the side AB shall be equal to the side AC. * 3. 1. + Hyp. For if AB be not equal to AC, one of them is greater than D. COROLLARY.-Hence, every equiangular triangle is also equilateral A 4. 1. B В с Q. E, D. PROPOSITION VII. See N. Theor.— Upon the same base, and on the same side of it, there cannot be two triangles that have their sides which are terminated in one extremity of the base, equal to one another, and likewise those which are terminated in the other extremily. A If it be possible, upon the same base AB, and upon the same side of it, let there be two triangles ACB, ADB, which have their sides CA, DA, terminated in the extremity A of the base, equal to one another, and likewise their sides CB, DB, that are terminated in B. Join CD; then, in the case in which the vertex of each of the triangles is without the other triangle, because AC is equal to AD, the angle ACD is equal* to the angle ADC: but the angle ACD is greater than the angle BCD; therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal + to DB, the angle BDC is equal to the angle BCD; but it has been demonstrated to be greater than it; which is impossible. But if one of the vertices, as D, be within the other triangle ACB, produce AC, AD to E, F : therefore, because, AC is equal + to AD in the triangle ACD, the angles ECD, FDC, upon the other side of the base CD, are equal * to one another : but the angle ECD is greater + than the angle BCD; wherefore + Нур. : * 5. 1. * Нур. : * 5. I. + 9 Ax. / ,E,F . 5. 1. the angle FDC is likewise greater than BCD; much more then is the angle BDC greater than the angle BCD. Again, because CB is equal to DB, the angle BDC + Hyp. is equal * to the angle BCD; but BDC has been proved to be greater than the same BCD; which is impossible. The case in which the vertex of one triangle is upon a side of the other, needs no demonstration. Therefore, upon the same base, and on the same side of it, there cannot be two triangles that have their sides, which are terminated in one extremity of the base, equal to another, and likewise those which are terminated in the other extremity. Q. E, D. A PROPOSITION VIII. THEOR.–1f two triangles have two sides of the one, equal to two sides of the other, each to each, and have likewise their bases equal; the angle which is contained by the two sides of the one, shall be equal to the angle contained by the two sides equal to them, of the other. Let ABC, DEF be two triangles, having the two sides AB, A D G For if the triangle ABC be applied to DEF, so that the point B may be on E, and the straight line BC upon EF, the point C shall also coin- B CE cide with the point F, because BC. is equal † to EF. Therefore BC coinciding with EF, BA | Hypand AC shall coincide with ED and DF: for if the base BC coincides with the base EF, but the sides BA, CA do not coincide with the sides ED, FD, but have a different situation, as EG, FG; the same base EF, and side of it, there can be two triangles that have their sides which are terminated in one extremity of the base, equal to one another, and likewise their sides terminated in the other extremity; but this is impossible; therefore, if the base BC * 7. 1. upon then upon the same coincides with the base EF, the sides BA, AC cannot but coincide with the sides ED, DF; wherefore likewise the angle BAC coincides with the angle EDF, and is equal * to it. Therefore, if two triangles, &c. * 8 Ax. Q. E. D. PROPOSITION IX. PROB.—To bisect a given rectilineal angle ; that is, lo divide it into two equal angles. Let BAC be the given rectilineal angle ; it is required to bisect it. Take any point D in AB, and from AC, cut off AE equal to AD; join DE, and upon it describe * an equilateral triangle DEF; then join AF: the straight line AF shall bisect the angle BAC. Because AD is equal t to AE, and AF is common to the two triangles DAF, EAF, the two sides DA, AF are equal to the two sides E EA, AF, each to each ; and the base DF is equal + to the base EF; therefore the angle B F DAF is equal* to the angle EAF: wherefore, the given rectilineal angle BAC is bisected by the straight line AF. Which was to be done. A + Constr. * 8.1. PROPOSITION X. Prov.-To bisect a given finite straight line; that is, to divide it into two equal parts. Let AB be the given straight line; it is required to divide it into two equal parts. Describe * upon it an equilateral triangle ABC, and bisect* the angle ACB, by the straight line CD: AB shall be cut into two cqual parts in the point D. D B Because AC is equalt to CB, and CD common to the two triangles ACD, BCD, the two sides AC, CD are equal to BC, CD, each to each ; and the angle ACD is equal + to the angle BCD; therefore the base AD is equal to the base * DB, and the straight line AB is divided into two equal parts in the point D. Which was to be done. A Constr. + Constr. * 4. I. PROPOSITION XI. PROB.- To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be a given straight line, and C a point given in it; See N. it is required to draw a straight line from the point C, at right angles to AB. Take any point D in AC, and make * CE equal to CD; * 3. 1. and upon DE, describe * the equilateral triangle DFE, and * 1. 1. join FC: the straight line Fc, drawn from the given point C, shall be at right angles to the given straight line AB. F Because DC is equal t to CE, and FC + Constr. common to the two triangles DCF, ECF, the two sides DC, CF are equal to the B two EC, CF, each to each ; and the base DF is equal t to the base EF; therefore the angle DCF is † Constr. equal to the angle ECF; and they are adjacent angles. 8. 1. But when the adjacent angles which one straight line makes with another straight line, are equal to one another, each of them is called a right * angle; therefore each of the angles • 10 Def. DCF, ECF, is a right angle. Wherefore, from the given point C, in the given straight line AB, FC has been drawn at right angles to AB. Which was to be done. CoR.-By help of this problem it may be demonstrated, that two straight lines cannot have a common segment. If it be possible, let the two straight lines ABC, ABD have the segment AB common to both of them. From the point B, drawt BE at right angles to AB; and because ABC is a +11. . straight line, the angle CBE is equal* to the . 10 Def. angle EBA; in the same manner, because E ABD is a straight line, the angle DBE is equal to the angle EBA ; wherefore + the angle DBE is equal to the angle CBE, the A B В с less to the greater; which is impossible: therefore, two straight lines cannot have a common segment. PROPOSITION XII. PROB.—To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point . without it. |