B А With D as centre, and at the distance D Е, describe a circle. (Post. III.) Produce the line DA to meet the circumference of this circle in the point F. (Post III.) Then A F will be a line E drawn as required. For The lines D E and DF are equal, being radü of the same circle. (Def. XIV.) The lines D C and D A are also equal, being sides of an equilateral A. If the equals, D C and D A, be taken from the equals, DE and DF, the remainders, C E and AF, will be equal. (Axiom III.) The lines, C B and CE, are equal, being radii of the same circle. So that the lines C B and AF are both equal to the line C E. Therefore C B and A Fare equal to each other. (Ax. I.) And as the line A F is drawn from the point A, and is equal to BC, it is drawn as was required. Having got one line drawn from the given point, and of the given length, we can draw a line of the same length from the same point in any other direction. Any line drawn from the point A to a point on the circumference of a circle drawn with A as a centre, and at the distance AF, will be of the same length as AF. Beginners always find this proposition very troublesome to remember. The principle upon which the construction is made must be carefully attended to. The given point may be joined with either extremity of the given line. The equilateral constructed on this joining line, may be constructed on either side of it. The circle described with the given line as a radiis, must have that end of the given line as a centre, which was joined by a right line to the given point. The centre with which the second circle is described must be that vertex of the equilateral A which is opposite to the line on which it was described. The radius with which the second circle is to be described is to be obtained by producing that side of the equilateral A which lies hetween this vertex and the end of the given line which was joined with the given point, until it cuts the circumference of the first circle. Then that side of the equilateral A which lies between the centre of the second circle and the given point, must be produced through the given point (not backwards through the centre of the second circle), till it meets the circumference of the second circle. As there are two ends to the given finite right line, with either of which the given point may be joined, and as the equilateral A may be constructed on either side of the joining line, there are four distinct constructions by which the problem may be solved in any given case. Thus, for the relative positions of the line and point given above, besides the construction depicted before, there are the three following: The letter B has been put where the letter C was, and C where B was, in the third and fourth cases, which allows the description of the construction, and the proof of its correctness, to remain word for word the same in all four cases. When the proposition has been mastered with the first construction, it should be carefully gone through with each of the other three ; and then all four methods should he applied to several cases where the given line and the given point are in different relative positions. By attending to the general directions, all possibility of mistake will be avoided. In repeating the proposition, let care be taken not to use the same letters to mark the different points of the figure. PROPOSITION III. From the greater of two given straight lines, to cut off a part equal to the less. To do this we must be able, equal to a given straight line. (Prop. II.) from that centre, to describe a circle. (Post. III.) To prove that the construction is correct, we must know,1. That radii of the same circle are equal. (Def.F.) 2. That magnitudes which are equal to the same are equal to each other. (Ax. I.) Let A B and C D be the two given straight lines, of which A B is the greater. From the point A draw the straight line A E equal to CD.* (Prop. II.) A B With A as centre, and at the distance A E, describe a circle, cutting the line A B in the * At this step the whole of the construction given in C E D point F. (Post. III.) A F is the part of A B which was required For, AE and AF are equal, being radii of the same circle (Def. XIV.); and as A E and C D are equal, A F and C D are both equal to A E. (Ax. I.) Therefore A F and C D are equal to one another; or, in other words, A F has been cut off from A B of the same length as C D. The shape of a is not determined if we know only the lengths of two of its sides. For an infinite number of As may be constructed, having two sides of the same length respectively. Thus, in the following As, the sides A B and BC are of the same length respectively in each : It is obvious, therefore, that we cannot pronounce two As to be of the same shape merely because two sides of the one are equal to two sides of the other, each to each ; for if we could, we should be able to say that all the above As were of the same shape. But if we also know the size of the < between A B and B C, then the shape of the A is determined. Let the above caution be carefully attended to. The neglect of it is a very common error with beginners. Prop. II. is supposed to be gone through, and the line A E to be a line obtained as the result of all the operations there indicated. Without going through the construction we cannot tell whether the line required will fall exactly in the position A E; but as the direction of the line A E is not a point of the least consequence for the rest of the construction, we need not trouble ourselves about it. PROPOSITION IV. If two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise the angle included be. tween those two sides in the one equal to the angle included between the corresponding sides of the other, then the bases, or third sides, will also be equal, and the remaining angles will be equal, each to each, those being equal to one another to which equal sides are opposite. In proving this proposition we are supposed to be able to place a given figure in any other position that we choose (see p. 9); and we are supposed to know 1. That two straight lines may be so placed one upon the other as to coincide with each other in direction, and cannut diverge from one another if they coincide for any portion of their length. (Ax. X.) 2. That two straight lines cannot enclose a space. (Ax. XI.) Suppose that A B C and DEF are two As, in which the sides A C and C B of the one are equal to the sides DF and FE of the other, each to each (i.e., AC to DF, and C B to F E), and the < ACB (included between the sides AC and C B) equal to the DFE (included between the sides D F and FE).* * Let it be observed, that this is all that we are able to assert about the As, to start with. Although it is true that all the other parts are also equal to each other, yet this has still to be proved, and before it has been proved we must on no account take it for granted, or speak of the equality of the other parts as a truth already known. F A B D E |