PROP. X. PROBLEM. To bisect a given finite straight line, i.e. to divide it into two equal parts. Let AB be the given finite straight line : It is required to bisect it. CONSTRUCTION.-1. On AB construct an equilateral triangle ABC (I. 1). 2. Bisect the angle ACB by CD, cutting AB in D (I. 9.) Then it is to be proved that The given straight line AB is bisected in D. PROOF.-Because in the triangles ACD and BCD we have the sides AC and CD, and their angle ACD, in the former the sides BC and CD, and their angle BCD, in the latter, each to each (cons.), therefore the base AD the base BD (I. 4). Therefore, it is proved, as required, that The given straight line AB is bisected in D. Q. E. F. Exercise. Given an isosceles triangle BAC with the vertical angle at A bisected by AD, meeting BC in D; prove that BC is bisected in D. PROP. XI. PROBLEM. To draw a straight line at right angles to a given Let AB be the given straight line, and C the given point in it: It is required to draw from C a straight line at right angles to AB. CONSTRUCTION. -1. Take any point D in AC. 2. Make CE = CD (I. 3). 3. Upon DE construct an equilateral triangle DFE (I. 1); 4. Join FC. Then it is to be proved that The straight line FC is drawn from C at right angles to AB. PROOF.-Because in the triangles DCF and ECF we have the three sides DC, CF, and FD, in the former = the three sides EC, CF, and FE, in the latter, each to each (cons.), therefore the angle DCF the angle ECF (I. 8), and these are adjacent angles, and therefore right angles (def. 10). Therefore, it is proved, as required, that The straight line FC is drawn from C at right angles to AB. Q. E. F. COROLLARY. By help of this Problem it may be demonstrated that Suppose it possible that ABC and ABD are two straight lines, with a common segment, or part, AB. CONSTRUCTION.-From B draw BE at right angles to AB (I. 11). PROOF.-Because ABC is a straight line (hyp.) with BE perp. to it (cons.), therefore the angle ABE = the angle EBC (def. 10); and because ABD is a straight line (hyp.) with BE perp. to it (cons.), therefore the angle ABE the angle EBD (def. 10); and therefore the angle EBD = the angle EBC (ax. 1), i.e. the less the greater, which is absurd (ax. 9). = The same result would follow for any other position of ABC and ABD, with AB as a part of each. Therefore, it is proved, as required, that Two straight lines cannot have a common segment. Q. E. D. Exercises. 1. Two points, A and B, are given above a straight line CD; find a point E in the straight line CD, so that if AE and BE be joined, AE = BE. 2. Find the same when A is above, and B below, the straight line CD, but the line joining A and B not perpendicular to CD. 3. Find the same when A is above, and B in, the straight line CD. PROP. XII. PROBLEM. To draw a straight line at right angles to a given straight line of unlimited length, from a given point without it. Let AB be the given straight line of unlimited length, or that which may be produced to any distance, both ways; and C the given point without it. It is required to draw from C a straight line at right angles to AB. CONSTRUCTION.-1. Take any point, D, upon the other side of AB. 2. From centre C with distance CD describe the circle EFG, cutting AB in F and G (post. 3). 3. Bisect FG in H (I. 10). 4. Join CF, CH, and CG. Then it is to be proved that The straight line CH is drawn from C at right angles to AB. PROOF.-Because in the triangles FHC and GHC, we have the three sides FH, HC, and CF, in the former = the three sides GH, HC, and CG, in the latter, each to each (cons.), therefore the angle FHC = the angle GHC (I. 8); and these are adjacent angles, and therefore right angles (def. 10). Therefore, it is proved, as required, that The straight line CII is drawn from C at right angles to AB. Exercises. Q. E. F. 1. Let ABCD be a square and AC and BD its diagonals; prove that AC = BD. 2. Let ABCD be a rhomboid; prove that its opposite angles each other. = 3. Let ABCD be a rhombus with diagonal AC; prove that the angle DAC = the angle BAC, and that the angle BCA: = the angle DCA. |