SUMMARY OF PROPOSITION VIII., THEOREM 5. If in two triangles the three sides of one are equal to the three sides of the other each to each, the triangles are equal in all respects. Cons.-Nil. Proof. By superposition. If the triangles do not coincide да The first group of theorems is now concluded; by their study we have added some important geometrical truths to those laid down in the axioms. These may be thus summed up : (1) In triangles equal angles are opposite to equal sides. Hence an isosceles triangle has two equal angles, and, conversely, a triangle with two equal angles is isosceles; an equilateral triangle is equiangular, and an equiangular triangle is equilateral (I. 5 and 6 cor.). (2) Triangles are equal in all respects, if in the triangles : (a) Two sides and the angle between them are equal each to each. (B) The three sides are equal each to each. D LECTURE VI. GEOMETRICAL ANALYSIS AND SYNTHESIS.-PROPS. 9-12.-TABLE OF RESULTS.-LOCI. IN the previous lessons we have not only been following Euclid's teaching, but it has been our aim to discover for ourselves the demonstrations of the theorems and problems proposed by him. In working out these we endeavoured to connect what we were required to prove with some other truths already established. One rule is often of great value in the investigation of a proposition, viz. to assume what we wish to prove, that is that the theorem is true or the problem solved, and to draw a diagram accordingly. The figure thus obtained will frequently suggest the lines by which the necessary construction may be effected. This method of proceeding is called analysis or resolution, and is pursued by geometers when they are required to find any new demonstration. After having worked out the proposition we retraced our steps, arranging and dividing it into its proper parts. This proceeding is termed synthesis, a Greek word, meaning a 'putting together,' an arrangement. Euclid's propositions are all examples of this method. Thus it will be seen that synthesis is the clearest method of stating a demonstration for the instruction of pupils or the conviction of opponents, while analysis is the means by which the demonstration is discovered. We will attempt the analysis of Prop. IX., availing ourselves of the rule just given. It is required to bisect a given angle—that is, to divide it into two equal angles. Draw any angle BAC and a line AX which, we will suppose, bisects it. Then angle BAX= angle CAX. To observe any results from the equality of AX is common to both, B E and the angles included by these lines are equal by hypothesis. Hence, if any angle is bisected, by joining equal arms of the angle with the bisecting line, equal triangles are formed. Can we reverse this order and construct equal triangles whose vertical angles shall together make up the angle BAC ? To do this we must fulfil the conditions either of I. 4 or of I. 8. As we have no angles given equal, we must follow I. 8 and endeavour to construct two triangles having each of their three sides equal. As before, by I. 3 we can make DA = EA. How can we obtain lines corresponding to DX and EX, i.e. two equal lines drawn from fixed points (D and E) and meeting at a point to be found? Since something is to be done, the postulates or the three problems we have learnt must help us. Examining the latter, we find that in Prop. I., 'to describe an equilateral triangle,' we were given two fixed A A D points, the extremities of the line AB, and from these we drew two equal lines meeting in a point C, not given. This exactly meets the present case, but before we can apply Prop. I. we require a base on which to describe the equilateral triangle. Glancing at the figure you will see that this can be obtained by joining D and E, the points from which the equal lines are to be drawn. Draw this line, and on it describe the equilateral triangle DFE (I. 1). The required triangles will be completed by joining FA, for in them: DA = EA, cons. DF = FE, def. 24. AF is common. Therefore their vertical angles DAF, FAE, are equal, and the given angle is bisected. SUMMARY OF PROPOSITION IX., PROBLEM 4. To bisect a given rectilineal angle. Cons.-Cut off from the arms of the given angle equal parts, AD, AE. Join DE. On it de scribe the equilat. A DFE. Join AF. Proof. The sides of the As DAF, EAF, are equal (I. 8). Therefore their verticals are equal, and BAC is bisected. [NOTE. In the construction of the equilateral triangle, the words 'on the side remote from A' have been inserted into the text of Euclid, because if the triangle were described on the same side as A, the points F and A might coincide, in which case the construction would fail; it holds good, however, if F falls either beyond or within A.] PROP. X.-The next problem is to 'bisect a given finite straight line.' Turning to the figure of the last proposition, you will observe that the base of the equilateral triangle is cut by the line AF at a point which we will call X, and that the equilateral triangle is divided by it into two equal triangles, DFX, EFX, for DF = EF (def. 24). FX is common. Angle DFX = angle EFX (I. 4). (Since they are corresponding angles of triangles DFA, EFA, shown to be equal.) Therefore the bases DX and EX are equal, and the line DE is bisected; and also since the angle DFX = EFX, the angle DFE is bisected. Again, the same line which bisects angle DFE bisects the line DE. C These considerations will enable us to solve the problem thus: on the given line describe an equilateral triangle (I. 1). Bisect its vertical angle (I. 9), producing the bisecting line to cut the base, i.e. the given line, which will then be itself bisected, as we have seen. SUMMARY OF PROPOSITION X., PROBLEM 5. Cons.-On given line describe an equilat. ▲ (I. 1). Bisect its vertical angle (I. 9), making the bisecting line cut the base. Proof.--The two As into which ABC is B divided are equal (I. 4); B D .. their bases are equal, and the given line is bisected. PROP. XI.-In the next problem it is required to draw a straight line at right angles to a given straight line from a given point in the same. From definition 10 we learn that when a straight line standing on another makes the adjacent angles equal, each of these angles is called a right angle.' Look at the diagram in the last proposition. Since the |