Propositions XXXIX. and XL. are the converse of XXXVII. and XXXVIII., and, as is usually the case, are proved by the indirect method. They show that equal triangles on the same base, and on equal bases in the same straight line, and on the same side of it, are between the same parallels. Let us take equal triangles ABC, DBC on the same base BC. Joining the vertices of the triangles, we are required to show that AD is parallel to BC. D Assuming that AD is not parallel to BC, draw AE parallel to BC. Looking at the figure you will see that AE will meet BD (or BD produced), so that by joining EC a triangle EBC will be formed between the same parallels AE, BC, and on the same base BC. Such triangles we know are equal (I. 37), i.e. triangle ABC triangle EBC; but triangle ABC is also = E equal to DBC, therefore EBC = DBC, the less to the greater, on the same base, &c., are between the same parallels. The construction and proof of Prop. XL. are precisely similar, making the reference to I. 38 instead of I. 37. SUMMARY OF PROPOSITION XXXIX., THEOREM 29. Equal triangles on the same base and on the same side of it are between the same parallels. Cons.-Draw AE || to BC. Join EC. Proof (Indirect).- ABC A EBC (I. 37). ▲ ABC = ▲ DBC (Hyp.). .. A EBC= A DBC, less to the greater, and AE is not || to BC, &c. SUMMARY OF PROPOSITION XL., THEOREM 30. Equal triangles on equal bases in the same straight line and on the same side of it are between the same parallels. Cons. and Proof same as above, making reference to I. 38 instead of I. 37. The results obtained from the propositions considered in this lesson may be briefly summed up. Parallelograms on the same base, or on equal bases in the same straight line and between the same parallels, are equal in area (I. 35 and 36). Under similar conditions triangles are equal in area (I. 39 and 40). And conversely, if equal triangles are on the same base or on equal bases in the same straight line, and on the same side of it, they are between the same parallels. If a parallelogram and a triangle are on the same base and between the same parallels, the triangle is half of the parallelogram (I. 41). LECTURE XVI. PROPOSITIONS 42-45. In this lesson, with the exception of one theorem, we shall be engaged in the consideration of some problems the solution of which depends on the principles which we have just learnt. The first is to describe a parallelogram that shall be equal to a given triangle and have one of its angles equal to a given angle. First we must ask what do we know about the relations of parallelograms and triangles in point of area? Prop. XLI. tells us that a parallelogram is double of any triangle on the same base and between the same parallels as itself. Then, if we can divide the given triangle into two equal triangles, we can construct a parallelogram double of either of these (as in I. 37) and consequently equal to the given triangle. But we have also to make one angle of the parallelogram equal to a given angle; this we can accomplish without difficulty by the use of I. 23, since the size of the angles will not affect the area of the parallelogram. One question remains, how are we to divide the given triangle into two equal triangles? We know from I. 37 and 38 that triangles are equal in area when they are between the same parallels and on the same or on equal bases. Let us endeavour to obtain these conditions. Let ABC be the given triangle, we have to divide it into two triangles on equal bases and between the same parallels. The equal bases can obviously be obtained by bisecting BC in E (I. 10). Then, if we draw a line through A parallel to BC, AK and BC will be the required parallels. Join EA, and F G D the required triangles are obtained. We have now to construct the parallelogram on one of the equal bases, say EC. First at the point E (I. 23) in line EC we will make an angle equal to the given angle D, and produce one of its arms to meet the opposite parallel AK in F. To complete the figure we require a side parallel to EF; from point C draw CG parallel to EF (I. 31), and meeting AK in G. solved, for FECG is by construction a opposite sides are parallel) having an angle equal to the given angle; also it is double of triangle AEC (I. 41), and therefore equal to triangle ABC. B E C The problem is thus parallelogram (i.e. its It follows from this proposition that any triangle may be divided into two others of equal area, by bisecting the base and joining the point of section with the vertical angle. SUMMARY OF PROPOSITION XLII., PROBLEM 11. To describe a parallelogram that shall be equal to a given triangle, and have one of its angles equal to a given rectilineal angle. Cons.-I. To bisect given triangle. Bisect BC in E. Join EA. Draw AK || to BC. II. To construct parallelogram. At E make CEF equal to D, let EF cut AK in F (I. 23). From C draw CG to EF, cutting AK in G (I. 31). Therefore AEC = ABC. D by cons. same parallels are equal (I. 38). FECG is gram having angle FECG is double of ▲ AEC—that is, of half ABC (I. 41). .. it is equal to ▲ ABC.-Q.E.D. I. 43. We pass on to a theorem which is necessary for the solution of the next problem. The complements of the parallelograms which are about the diameter of any parallelogram are equal to one another.' A H. D The parallelograms about the diameter' of any parallelogram are those whose diameters consist of a portion of that of the original parallelogram. The complements are those portions of the figure which, together with the parallelograms about the diameter, make up the whole parallelogram. Draw a parallelogram of which AC is the diameter, then taking any point K in AC draw parallelograms of which AK and KC are diameters; these are parallelograms about AC. Together with these, two small parallelograms, EKGB and HKFD, make up the whole original parallelogram ABCD; these are called the complements, since they complete or fill up the figure, and what the theorem asserts is, that these complements are equal in area. And that this is true you will easily see. large parallelogram and 'the parallelograms about the diameter' are each divided by their respective diameters into two triangles, which we know to be equal (I. 34). Also each large triangle is made up of one of each of the smaller triangles together with one complement. Then, taking the smaller triangles which are equal from the large triangles which are equal, it is clear that the remainders, the complements, are equal. K E F For the B G There may be three cases in the figure of this theorem, but the same proof applies to each. For the diameters of the 'parallelograms about the diameter' may be together equal to the diameter of the original parallelogram, as in the example we have taken, (2) or they may only form a part of it, (3) or the parallelograms may overlap so that their diameters are together more than the diameter of the original parallelogram. In the second case the complements are trapeziums, not parallelograms. H |