from AB for the Triangle, he will make use of the Flat-ruler with cross line, precisely as in (102). (4) The Carpenter and Mason will apply the 'Square' in a similar manner. 104. PROP. VII. Through a given point to draw a straight line which shall be parallel to a given straight line. (1) This is done in two ways in (36); and the latter method is sufficiently practical with no other instrument than the ordinary Square', or 'Triangle'. (2) But the same thing is most readily done by means of the Parallel-Ruler', made for the purpose, unless the given point be at a greater distance from the given line than the extreme width of the ruler when opened to its fullest extent. If the given point be nearer to the line than the width of the ruler when closed, place the whole ruler on the other side of the given line with one edge exactly along the line; then move this edge, while the other half of the ruler is held tight, until it exactly passes through the given point, and draw the required line along this edge in that position. If the point be at a distance from the line somewhat greater than the width of the ruler when closed, place the closed ruler between the point and the line, with one outer edge coinciding with the line; hold this tight, and move the other outer edge until it passes through the point; then draw along it the line required. 6 (3) Another method is by means of the Triangle" and Flat-ruler'. Let AB be the given straight line, and C the given point. Place the Triangle on the opposite side of AB to C, so as to have one of the sides forming the right angle exactly on AB. Then lay the Ruler along the hypothenuse of the Triangle, as DE in the annexed fig., and while the Ruler is held tight in that position, slide the Triangle along it, until the same side which was on AB passes through C. Then draw FCG along that side, and it shall be parallel to AB, and may be produced both ways from F and G as required. FCG is parallel to AB, because they are two straight lines intersected by another straight line DE, making the exterior angle AHD equal to the interior and opposite angle FGH on the same side of it (34 Cor. 1). (4) Another Method. With centre B, any point in the given line distant from A, and radius BC, describe the arc CD cutting AB in D. With centre D, and the same radius as before, describe the arc BE on the same side A D B of AB; and from BE measure off, with the compasses or otherwise, BF equal to CD, remembering that in circles of the same radius equal chords subtend equal arcs (58). Join CF, and it is parallel to AB. For CD, BF being equal arcs of equal circles, they subtend equal angles at the centre (59), that is, the alternate angles CBD, BCF are equal, CF is parallel to AB (34). From the point C draw any From (5) Another Method. straight line CE meeting D; and make DE = CD. E draw another line EG cutting AB in F, and make FG = EF. Join CG, and CG is the line required. For the two sides EC, EG of the triangle ECG are divided pro C G portionally in the points D and F, .. CG is parallel to DF (70). (6) The Carpenter or Mason will employ a method still more simple: By means of his Square' he will draw CD perpendicular to AB; and from any other point E in AB he will draw EF at right angles to AB, making EF CD. Then join CF, and CF is parallel to AB (35 Cor.) = A D E It is upon this principle of equidistance of parallel lines that the instrument called a Joiners' Gauge is constructed. 105. PROP. VIII. From a given point in a given straight line to draw another straight line making with the former an angle equal to a given angle. E (1) Let AB be the given straight line; C the given point in it; and EDF the given angle. (The angle is here supposed to be given by being traced on the same plane as that on which the other angle is to be drawn.) With centre D and any convenient radius, the greater the better, describe the arc EF. With centre C, and the same radius, describe the arc GH. Then A B with centre G and radius equal to the chord EF, describe an arc cutting GH in H. Join CH, and it shall be the straight line required. For, if EF, GH be joined, they are equal chords of equal circles,.. the arcs EF, GH are equal, and .. GCH = LEDF (59). (2) If either of the lines which form the given angle, EDF, be in the same straight line with, or parallel to, the given line, AB, then it will only be necessary to draw through C, (by means of the Parallel-Ruler or otherwise), a straight line parallel to the other side of the angle EDF. 106. PROP. IX. Through a given point without given straight line to draw another straight line making with the former an angle equal to a given angle. E D Let AB be the given straight line, and C the giver point. Through C draw CD parallel to AB. From C draw CE making with CD the angle DCE equal to the given angle (105). Produce EC to meet AB in F, and EF is the line required. For the straight line EF meets the parallel lines AB, CD; :. ▲ BFC = ▲ DCE=the given angle (34). A F B 107. PROP. X. To construct an equilateral triangle having each of its sides equal to a given straight line. This is done in (23); and no better method can be adopted in practice. 108. PROP. XI. To construct a triangle with its sides respectively equal to three given straight lines. Let A, B, C, be the three given straight lines. Take DE equal to A. With centre Dand radius equal to B describe of DE on which the triangle is A B C D EF; and DEF is plainly the triangle required. N.B. Since in every triangle any two sides are together greater than the third side (38), the given straight lines must be such that any two of them are greater than the third. COR. If A=B = C, the same construction will hold, and the triangle is equilateral. If B=C, the triangle is isosceles. 109. PROP. XII. To construct a triangle with two sides respectively equal to two given straight lines and an angle equal to a given angle. Let A, B, be the given straight lines, and C the given angle. (1) In the case when the angle C is to be between the given sides, draw DE=A; at the point D make 4 EDF=4 C (105), and DF-B. Join EF, and DEF is the triangle required. (2) In the case when the angle C is to be opposite to one of the given sides, as B, proceed as before, but instead of cutting off DF equal to B, with centre E A B and radius B describe an arc cutting DF in F. Then join EF, and DEF is the triangle required. 110. PROP. XIII. To construct a triangle with two angles respectively equal to two given angles, and one side equal to a given straight line. Let A be the given straight line, and B, C, the two given angles. (1) In the case when the given side is to be adjacent to both the given angles, take DE=A ; make ‹ EDF=B (105), A and ▲ DEF=C. Then DEF is the triangle required. B F D D که E E OBS. In every triangle there are six parts, three sides and three angles; and of these six if any three be given, except three angles, the triangle is determined. But, it is plain that a triangle is not known, when its angles only are known, because an infinite number of different triangles may have the same or equal angles, as will easily appear by drawing within any given triangle straight lines parallel to the sides. PART II. 2 |