the rectangle CB, BD; that is, the square on AC alone is less than the squares on CB, BA by twice the rectangle CB, BD. Next, let the side AC be perpen A dicular to BC. Then BC is the straight line between the perpendicular and the acute angle at B; and it is manifest, that the squares on AB, BC are equal to the square on AC, and twice the square on BC. [I. 47 and Ax. 2.] Therefore, in every triangle, &c. Q.E.D. This and the previous proposition should be compared with I. 47. Taken together they show that the squares on two sides of a triangle are together equal to greater than or less than the square on the third side according as the angle contained by these two sides is right, obtuse or acute. Ex. Drop a perpendicular from B upon AC meeting AC in E, and show that the rectangle AC, CE is cqual to the rectangle BC, CD. PROPOSITION 14. PROBLEM. To describe a square that shall be equal to a given rectilineal figure. Let A be the given rectilineal figure: it is re1. quired to describe a square that shall be equal to A. Describe the rectangular parallelogram BCDE 2. equal to the rectilineal figure 4. Then if the sides of it, BE, ED are equal to one another, it is a square, and what was required is now done. But if they are not equal, produce one of them BE to F, and make EF equal to ED, and bisect BF in G; [I. 3, I. 10.] from the centre G, at the distance GB, or GF, describe the semicircle BHF, and produce DE to H. The square described on EH shall be equal to the given rectilineal figure A. Join GI. Then, because the straight line BF is divided 3. into two equal parts at the point G, and into two unequal parts at the point E, the rectangle BE EF, together with the square on GE, is equal to the square on GF. [II. 5.] But GF is equal to GH. Therefore the rectangle BE, EF, together with the square on GE, is equal to the square on GH. But the square on GH is equal to the squares on GE, EH; [I. 47.] therefore the rectangle BE, EF, together with the square on GE, is equal to the squares on GE, EH. Take away the square on GE, which is common to both; therefore the rectangle BE, EF is equal to the square on EH. [Ax. 3:] But the rectangle contained by BE, EF is the parallelogram BD, because EF is equal to ED. [Const.] Therefore BD is equal to the square on EH. But BD is equal to the rectilineal figure A. [Const.] Therefore the square on EH is equal to the rectilineal figure A. Therefore a square has been made equal to the given rectilineal figure A, namely, the square described on EH. Q.E.F. From this proposition it appears that if a perpendicular EH be drawn from any point of a semicircle to the diameter, the square on this perpendicular is equal to the rectangle contained by the segments BE, EF into which it divides the diameter. Ex. To cut a line so that the rectangle contained by its segments shall be equal to the square on a given straight line. EXERCISES. From the truths established in the foregoing propositions other truths may be deduced in addition to those alrready proved. A few such examples have already been given as riders to the propositions, and before introducing more difficult exercises a few directions and illustrations will be given for the sake of shewing how we are to proceed in the demonstration of new propositions. In every proposition then there are two parts to be considered, 1st, the truths given ; 2nd, the truths to be proved. In investigating a new proposition, therefore, we always commence by drawing out and stating in words the truths given in the hypothesis, and applying them to a particular figure. Starting with these truths we desire to arrive at a certain result, and the next step is therefore to set clearly before the mind the truth to be proved. The question now is, What truths will connect these two things? How shall we, commencing with the hypothesis, deduce other truths which shall lead us to the desired conclusion? No answer can be given to this question which would be applicable in every case; but the following illustration will suggest the proper course. Suppose, for instance, we are required to prove that, under given conditions stated in the hypothesis, two lines are equal to one another. Then if these lines are sides of two triangles, or by a suitable construction may be made to become so, we consider how these two triangles may have their sides proved equal. But the propositions which enable us to prove the sides of triangles equal are the 4th and 26th, and looking back to the hypothesis we soon see which of these is applicable in the given case. If two sides of the triangles are given equal by the hypothesis we apply the 4th proposition, but if two angles, the 26th. This mode of procedure may be applied to the following exercise. If two angles of a triangle be bisected the three perpendiculars dropped upon the sides of the triangle from the points of intersection of these bisecting lines will be equal to one another. [The figure may easily be supplied.] Let ABC be the triangle, and BO, OC straight lines bisecting the angles B and C and intersecting in 0. From O drop the perpendiculars OD, OE, OF upon the sides BC, AC and AB respectively, then these straight lines shall be equal to one another. If now the hypothesis be gone over and applied to this fignre it will be seen that the angles FBO, OFB in the triangle OPB are equal to DBO, BDO in the triangle OBD and similarly with the triangles ODC, OEC. Next, looking to the result required, we see that OF and OD, which are to be proved equal to one another, are sides of the triangles OFB. ODB. A very little consideration will shew that since OB is common, the 26th proposition will connect the hypothesis with the conclusion, and the proof should be proceeded with accordingly. Any result involving the equality of the sides and angles of triangles would generally be attained in a similar way by reference to propositions 4, 8, and 26 ; but which of them is to be used in any given case can only be determined by comparing the hypothesis of the exercise with the hypothesis of each proposition. On the other hand, if the sides or angles to be proved equal form parts of the same triangle the propositions to be used will be the 5th or 6th, and so on with respect to other cases. In the first set of exercises which follow the propostions to be used are suggested by the headings of each group which form an analysis of the principal contents of Book I. GEOMETRICAL ANALYSIS. It has been already pointed out that the general course of procedure in Euclid's propositions is to begin with some truth known or admitted, and from this to deduce some new truth. This method is termed the 'synthetical method. But according to the 'analytical method' we begin by assuming the truth of the proposition. We then from this assumption deduce other results which we can compare with truths already known. If the results thus deduced do not agree with truths already known we conclude that the theorem is false or that the problem is insoluble ; but if they agree with truths already known, then setting out with these results, we may generally retrace our steps and from them deduce the proposition we are required to prove. True this cannot always be done, but the method is of such wide application that an account of it cannot be altogether omitted, and although it has been already illustrated in part, a more particular account of its application to the solution of problems will now be given. 1. Through a given point to draw a straight line equally inclined to two given straight lines which cut one another. Let AB, AC be the two given lines and P the given point. Suppose POR the required line drawn through P, so that the angles ARO and AOR are equal. Bisect RO in D and join AD. Then the triangles ARD, AOD have the three sides of the one equal to the three sides of the other, and therefore by I. 8 we can prove the angles at A equal to one another and likewise the angles at D. Hence the angles at D are right angles, and AD bisects the angle BAC. iP A The following is therefore the synthetical construction. Bisect the angle BAC between the lines, and from P the given point drop PD perpendicular to the bisecting line and meeting AB, AC in R and O. POR is the line required. 2. To bisect a quadrilateral by a straight line drawn through one of its angular points. Let ABCD be the quadrilateral, and suppose AE drawn through the angle A bisects the figure. Draw the diagonals B AC, BD and bisect BD A. in 0. Join AO, CO. Then the triangles AOD, AOB are equal I 38, as also the triangles BOC, DOC. Hence AOD and DOC are together equal to AOB, COB, and therefore the broken line AOC bisects the quadrilateral. But AEC is half the D quadrilateral, therefore AECB=OECB. Take away the common part ABC, then the triangle AEC is equal to the triangle AOC; and they are on the same base, therefore they are between the same parallels and EO is parallel AC. Hence the following construction. Draw the diagonals DB and AC. Bisect DB the diagonal which does not pass through the given angular point in 0. Through O draw OE parallel to the other diagonal and join AE which shall bisect the quadrilateral. The method of substituting a straight line for a broken one here employed should be carefully noted as it has many applications. |