and the angle DBC is equal to the angle ACB; (hyp.) therefore the base DC is equal to the base AB, (1. 4.) and the triangle DBC is equal to the triangle ACB, the less equal to the greater, which is absurd. Therefore AB is not unequal to AC, that is, AB is equal to AC. COR. Hence every equiangular triangle is also equilateral. PROPOSITION VII. THEOREM. 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, on 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, DŽ, that are terminated in B. First. When the vertex of each of the triangles is without the other triangle. Because AC is equal to AD in the triangle ACD, therefore the angle ACD is equal to the angle ADC; (1.5.) but the angle ACD is greater than the angle BCD; (ax. 9.) therefore also the angle ADC is greater than BCD; much more therefore is the angle BDC greater than BCD. Again, because the side BC is equal to BD in the triangle BCD, (hyp.) therefore the angle BDC is equal to the angle BCD; (1.5.) but the angle BDC was proved greater than the angle BCD, hence the angle BDC is both equal to, and greater than the angle BCD; which is impossible. Secondly. Let the vertex D of the triangle ADB fall within the triangle AČB. E Produce AC and AD to E and F. Then because AC is equal to AD in the triangle ACD, therefore the angles ECD, FDC upon the other side of the base CD are equal to one another; (1. 5.) but the angle ECD is greater than the angle BCD; (ax. 10.) much more then is the angle BDC greater than the angle BCD. Thirdly. 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, &c. Q.E.D. 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 angles 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. Then the angle BAC shall be equal to the angle EDF. For, if the triangle ABC be applied to DEF, so that the point B be on E, and the straight line BC on EF; then because BC is equal to EF, (hyp.) therefore the point C shall coincide with the point F; BA and AC shall coincide with ED, DF; for, if the base BC coincide with the base EF, but the sides BA, AC do not coincide with the sides ED, DF, but have a different situation as EG, FG: Then, upon the same base, and upon the same side of it, there can be two triangles which have their sides which are terminated in one extremity of the base equal to one another, and likewise those sides which are terminated in the other extremity; but this is impossible. (1. 7.) Therefore, if the base BC coincide 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, (ax. 8.) Therefore if two triangles have two sides, &c. Q. E.D. To bisect a given rectilineal angle, that is, to divide it into two equal angles. Take any point D in AB; from AC cut off AE equal to AD; (1. 3.) and join DE, describe an equilateral triangle DEF on the side of DE remote from A, (1. 1.) and join AF. Then the straight line AF shall bisect the angle BAC. 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 the base EF; (constr.) therefore the angle DAF is equal to the angle EAF. (1.8.) Wherefore the angle BAC is bisected by the straight line AF. Q. E. F. PROPOSITION X. PROBLEM. 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 AB into two equal parts. D Upon AB describe the equilateral triangle ABC; (1. 1.) and bisect the angle ACB by the straight line CD meeting AB in the point D. (1.9.) Then AB shall be cut into two equal parts in the point D. Because AC is equal to CB, (constr.) and CD is 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; (constr.) therefore the base AD is equal to the base DB. (1. 4.) Wherefore the straight line AB is divided into two equal parts in the point D. Q.E. F. PROPOSITION XI. PROBLEM. 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 given point in it. It is required to draw a straight line from the point C at right angles to AB. In AC take any point D, and make CE equal to CD; (1. 3.) upon DE describe the equilateral triangle DEF, (1. 1.), and join CF. Then CF drawn from the point C shall be at right angles to AB. Because DC is equal to CE, and FC is common to the two triangles DCF, ECF; the two sides DC, CF are equal to the two sides EC, CF, each to each; and the base DF is equal to the base EF; (constr.) therefore the angle DCF is equal to the angle ECF: (1.8.) and these two angles are adjacent angles. But when the two adjacent angles which one straight line makes with another straight line are equal to one another, each of them is called a right angle: (def. 10.) 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. Q.E.F. 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 segment AB be common to the two straight lines ABC, ABD. From the point B, draw BE at right angles to AB; (1. 11.) therefore the angle ABE is equal to the angle EBC; (def. 10.) but the angle ABE is equal to the angle EBC, PROPOSITION XII. PROBLEM. 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 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 F and G; (post. 3.) bisect FG in H (1. 10.), and join CH. Then the straight line CH drawn from the given point C shall be perpendicular to the given straight line AB. Join CF, and CG. And because FH is equal to HG (constr.), and HC is common to the triangles FHC, GHC; the two sides FH, HC, are equal to the two GH, HC, each to each; and the base CF is equal to the base CG; (def. 15.) therefore the angle FHC is equal to the angle GHC; (1.8.) But when a straight line standing on another straight line, makes the adjacent angles equal to one another, each of them is a right angle, and the straight line which stands upon the other is called a perpendicular to it. (def. 10.) Therefore from the given point C a perpendicular CH has been drawn to the given straight line AB. Q. E. F. PROPOSITION XIII. THEOREM, The angles which one straight line makes with another upon one side of it, are either two right angles, or are together equal to two right angles. Let the straight line AB make with CD, upon one side of it, the angles CBA, ABD. Then these shall be either two right angles, or shall be together, equal to two right angles. For if the angle CBA be equal to the angle ABD, But if the angle CBA be not equal to the angle ABD, therefore the angles CBE, EBD are equal to the three angles CBA, ABE, EBD. (ax. 2.) |