common to groups A and B; therefore BE+ EC are less than BA + AC; and since we have shown BD + DC to be less than BE + EC, they must also be less than BA + AC. II. To show that the angle BDC is greater than the angle BAC. E We know that one angle may be shown to be greater than another if in any triangle it has the greater side opposite (I. 19); or, if the two angles are not in the same triangle, it can be shown to be the exterior angle of a triangle, of which the other is an interior and opposite angle (I. 16). The latter condition will be the most likely to help us here. amining the figure you will see Ex B angle BDC is greater than ▲ BEC (I. 16). You will observe that in this theorem, as in one or two previous, we were unable to compare the lines and angles in question directly with each other, but a third set of lines and a third angle became, as it were, our medium of comparison. This method is often very useful in mathematical demonstrations. SUMMARY OF PROPOSITION XXI., THEOREM 14. If from the ends of the base there be drawn two straight lines to a point within a triangle, these lines will be less than the other two sides of the triangle, but will contain a greater angle. Cons.-Produce BD to meet AC in E. Proof.-I. To show that BD, DC are less than BA, AC. DE, EC are greater than DC (I. 20), add BD. Then BE, EC are greater than BD, DC. B D E Similarly BA, AC are greater than BE, EC, and consequently than BD, DC. II. To show that angle BDC is greater than angle BAC. Ext. BDC is greater than int. and opposite / DEC (I. 16). Ext.DEC is greater than int. and opposite ▲ BAC. :. ▲ BDC is greater than ▲ BAC. The group of theorems which we have just concluded extends from Props. 13-21, and may be divided into two sections. (a) Props. 13, 14, 15, which treat of the angles made by straight lines either meeting or cutting each other. The results of these have been already summed up (v. p. 45). (B) Props. 16-21 inclusive, treating of various properties of triangles; the results obtained in these we shall discuss a little later on. LECTURE X. PROPOSITIONS 22-23. I. 22. Two problems follow. Prop. 22 requires us to construct a triangle, of which the sides shall be equal to three given straight lines, but any two of these must be greater than the third.' Let A, B, C represent the three given lines, then, in order that we may have a connected measure for these lines, let us mark off on a line of unlimited length DF = line A, FGB, and GH = C, then for the base of the triangle we can take any one of these portions. You will remember that in the very D first problem we G H had to deal with, we were required to construct a triangle (equilateral); the base was then given us, and we were able to make the other two sides by drawing circles which cut each other. Can we apply this method here? Yes, for having chosen our base, we can take the two remaining portions as radii of two circles. But in order to make a triangle it will be necessary that these circles should cut as in the figure Similarly BA, AC are greater than BE, EC, and consequently than BD, DC. II. To show that angle BDC is greater than angle BAC. Ext. BDC is greater than int. and opposite DEC (I. 16). Ext. DEC is greater than int. and opposite BAC. .. BDC is greater than ▲ BAC. LECTURE X. PROPOSITIONS 22-23. I. 22. Two problems follow. Prop. 22 requires us to construct a triangle, of which the sides shall be equal to three given straight lines, but any two of these must be greater than the third.' Let A, B, C represent the three given lines, then, in order that we may have a connected measure for these lines, let us mark off on a line of unlimited length DF = line A, FGB, and GH = C, then for the base of the triangle we can take any one of these portions. You will remember that in the very D first problem we F G H E |