PROPOSITION IX. PROBLEM. To bisect a given rectilineal angle, that is, to divide it into two equal parts. Let BAC be the given rectlin. L. It is required to bisect it. Constr. Take any pt. D in AB; from AC cut off AE = AD, (i. 3) and join DE, on the side of DE remote from A, describe an equilat. A DEF, (i. 1) and join AF. Then AF shall bisect _BAC. Dem. : AD= AE (constr.), and AF is common to the As DAF, EAF; the two sides AD, AF = the two sides EA, AF, each to each; and base DF=base EF, (def. 24) .: LDAF= LEAF. (i. 8) Wherefore, the angle BAC is bisected by the straight line AF Q. E. F. [The learner should here, and in all following problems, find out for himself the data and quæsita. The equilat. A is described on the side remote from A because the three lines DA, AE, EF might possibly be all equal, that is DAE might itself be an equilat. A. In this case if the equilat. A DFE were described on the same side of DE on which A is, the vertex F would fall exactly on A, and there would be no line AF. Also if the <BAC were such that it would fall within the equilat. A DEF when described on the same side of DE as A, the line AF would not in this case bisect Z BAC, and if we produced FA to some pt. G we do not at present know that ZDAG= ZEAG: this is proved by prop. 13. Observe that by help of this prop. an may be divided into 4, 8, 16 &c. equal Zs.] PROPOSITION X. PROBLEM. To bisect a given finite straight line, that is, to divide it into two equal parts. Let AB be the given finite st. line. Constr. On AB describe the equilat. A ABC; (i. 1) and bisect LACB by CD meeting AB at D. (i. 9) Then AB shall be bisected at D. Dem. : AC = BC, (def. 24) and CD is common to As ACD, BCD; the two sides AC, CD= BC, CD each to each ; and LACD= _ BCD; (constr.) .. base AD= base DB (i. 4) Wherefore the straight line, &c. 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 st. line, and C the given pt. in it. Constr. In AC take any pt. D, and make CE = CD; (i. 3) On DE describe the equilat. A DEF, (i. 1) and join CF, Then CF drawn from the pt. C shall be I AB. Dem. •: DC= CE, (constr.) and FC is common to As DCF, ECF; .. the two sides DC, CF = the two sides EC, CF, each to each ; and the base DF = the base EF, (def. 24) .. _ DCF = _ECF; (i. 8) and these two Ls are adj. Ls. But when a st. line standing on another st. line makes the adj. Ls equal to one another, each of these Zs is a rt. L: (def. 10) . each of Zs DCF, ECF is a rt. L. Therefore, from the given pt. C in the given st. line AB, CF has been drawn at right angles to AB. Q. E. F. Cor. By help of this problem, it may be demonstrated that two st. lines cannot have a common segment. If possible, let the segment AB be common to the two st. lines ABC, ABD. E D B From B draw BE I AB. Then : ABC is a st. line, (hyp.): LABE = L EBC; (def. 10) Similarly : ABD is a st. line, (hyp.):: L ABE= EBD; but _ ABE = _ EBC, .. LEBD= EBC; (ax. 1) the less < = the greater L, which is impossible. .:. two st. lines cannot have a common segment. [Avoid a very common improper application of prop. 8. After saying, and the base DF = the base EF,' boys often go on ... the As DCF, FCE are equal in all respects,-by i. 8,' &c. Now Euclid does not prove this in prop. 8, and it is an error to apply that prop. in this manner. This remark might be repeated in every prop. in which prop. 8 is referred to. The proof of the corollary is unsound, because the prop. does not tell us how to draw a perpendicular to a line from its extremity. It might come after prop. 13: the proof is left as an exercise for the student. It would however perhaps be better to class the theorem that two st. lines cannot have a common segment' with the axioms, since it seems to proceed from the definitions, and it is tacitly assumed in prop. 1; (viz. that AC, BC, cannot have a common segment at C).] Describe a circle which shall pass through two given points and have its centre in a given straight line Is this always possible? |