А B D E two sides FA, AC, are equal to the two GA, AB, each to each; and they contain the angle FAG common to the two triangles AFC, AGB; therefore the base FC is equal to the 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., ACF to ABG, and AFC to 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 to be equal to GB; therefore the two BF, FC, are equal to the two CG, GB, each to each, and the angle BFC is equal to CGB, and BC is common to the two triangles BFC, CGB; wherefore the triangles are equal and their remaining angles each to each, to which the equal sides are opposite; therefore FBC is equal to GCB, and BCF to CBG: and, since it has been demonstrated that the whole angle A BG is equal to the whole ACF; the parts of which, CBG, BCF, are also equal, the remaining angle ABC is therefore equal to the remaining angle ACB; which are the angles at the base of the triangle ABC; and it also has 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. COROLLARY.—Hence every equilateral triangle is also equiangular. : VI.-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 triar.gle having the angle ABC equal to the angle ABC; the side AB is also equal to the side AC. For if AB be not equal to AC, one of them is greater: let AB be the greater and from it cut off A D B с BD equal to AC the less, and join DC; therefore, because in the triangles DBC, ACB, DB is equal to AC, and BC common to both, the two sides DB, BC, are equal to the two AC, CB, each to each; and the angle DBC is equal to ACB; therefore DC is equal to AB, and the triangle DBC is equal to the triangle ACB, the less to the greater which is absurd. Therefore AB is not unequal to AC, that is, it is equal to it. COROLLARY.—Hence every equiangular triangle is also equilateral. VII.—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 extremity. If it be possible, let there be two triangles ACB, the same base AB, and upon the same sido of it, which have their sides CA, DA, terminated in the extremity A of the baso 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, because AC is equal to AD, the angle ACD is equal to ADC; but the angle ACD is greater than BCD; Á therefore the angle ADC is greater also than BCD; much more then is the angle BDC greater than BCD. Again, because CB is equal to DB, the angle EDC is equal to BCD; but it has been demonstrated to be greater, which is impossible. But if one vertex, as D, be within the other triangle ACB, produce AC, AD, to Eand 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; but the angle ECD is greater than BCD; wherefore the angle FDC is likewise greater than ADB, upon B F BCT); much more then is the angle BDC greater than BCD. Again, because CB is equal to E DB, the angle BDC is equal to 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. B D G CE VIII.-If 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, AC, equal to the two sides DE, DF, each to each, viz., AB to DE, and AC to DF: and also the base BC equal to the base EF. The angle BAC is equal to the angle EDF. For, if the triangle ABC be 3 applied to DEF, so that the point B be on E, and the straight line BC_upon EF; C shall also coincide with F, because BC is equal to EF. Therefore BC coinciding with EF, BA and 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, then upon the same base EF, and on the same 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 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. D F IX.—To bisect a given rectilineal angle, that is, divido 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 A E, equal to AD; join DE, and on it describe an equilateral triangle DEF; join AF; AF bisects the angle BAC. Because AD is equal to AE, and AF is common to the two triangles DAF, EAF; the two sides DA, AF, are equal to the two sides EA, AF, each to each; and the base DF is equal to EF; therefore the angle DAF is equal to EAF: B wherefore the given rectilineal angle BAC is bisected by the straight line AF. X.-To bisect a given finite straight line, that is, divide it into two equal parts. Let AB be the given straight line; it is required to divide it into two equal parts. Describe on it an equilateral triangle ABC, and bisect the angle ACB by the straight line CD. AB is cut into two equal parts in the point D. Because AC is equal to CB, and CD A 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 BCD; therefore the base AD is equal to DB, and the straight line AB is divided into two equal parts in the point D. XI.—To draw a straight line at right angles to a given straight line, from a given point in the same. Let AB be the given straight line, and C a point given in it; it is required to draw a straight line from the point C at right angles to AB. Take any point Din AB, and make CE equal to CD, and upon DE describe the equilateral triangle DFE, and join FC. FC, drawn from the given point C, is at right angles to AD the given straight line AB. D B F E B Because DC is equal to CE, and FC common to the two triangles DCF, ECF; the two sides DC, CF, are equal to the two EC, CF, each to each; and the base DF is equal to the base EF; therefore the angle DCF is equal to ECF; and they are adjacent angles. 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 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. COROLLARY.-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 draw BE at right angles to AB; and because ABC is a straight line, the angle CBE is equal to the angle EBA; in the same manner, because 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 less to the greater, which is impossible; therefore two straight lines cannot have a common segment. D A B XII.—To draw a straight line perpendicular to a given straight line of an unlimited length, from a given point without it. Let AB be the given straight line, which may be produced to any length both ways, and let C be a point without it. It is required to draw a straight line perpendicular to AB from the point C. Take any point D upon the other side of AB, and from the centre C, at the distance CD, describe the circle EGF meeting AB in FG; and bisect FG in H, and join CF, CH, CG; the H GB D |