LECTURE XI. PROPOSITIONS 24-26. THE next two theorems teach us another relation between the sides and angles of triangles; i.e. the relation between the base and the vertical angle of a triangle. Supposing the two sides of a triangle to be of a certain length, the greater the vertical angle is, the greater the base will be, and conversely. I. 24. Euclid enunciates the first of these theorems thus: If two triangles have two sides of the one equal to two sides of the other each to each, but the angle contained by the two sides of the one (i.e. the vertical angle) greater than the angle contained by the two sides equal to them of the other, the base of that which has the greater angle shall be greater than the base of the other. = Let us take two triangles ABC, DEF, having their sides BA, AC to the sides DE, DF, but let the angle BAC be greater than the angle EDF, we have to show that the base BC is greater than the base EF. The first thing that will strike you is, that in order to compare these triangles we must connect them in some way; in the 4th and 8th props., in which two triangles are compared with each other, we employed the method of superposition. Now instead of supposing one triangle to be taken up and placed on the other, a method which we have seen is sparingly used by Euclid, by the help of the last two props. we can construct a triangle equal to ABC and having an angle EDG equal to the angle BAC, and as DE and AB are equal, we can make DE one side of the new triangle, drawing DG equal to AC, and since the angle EDG= angle BAC, the base EG will be equal to BC. So that the triangle EDG may be taken to represent the triangle ABC, and we have to show that EG is greater than EF. да E G F In drawing the figure we have just described you will find that the point F may fall on EG, so that the base E F forms a part of EG, or it may fall without the triangle EDG so that EF is below EG, or lastly it may fall within triangle EDG so that EF is above EG. (See fig. below.) The case in which F falls on EG needs no demonstration, for EG is necessarily greater than its part EF. Let us consider the second case in which EF falls below EG, then we may connect EF and EG in one triangle by joining FG. And we know that the greater side of any triangle is opposite the greater angle (I. 19). Is angle EFG greater than angle EGF ? : Examining the angles we find that :angle EFG is greater than DFG (its part) DFGDGF (I. 5) DGF is greater than EGF. Therefore angle EFG is greater than EGF, and EG (i.e. BC) is greater than EF. Lastly we have the third case in which DF falls above DG, joining FG a triangle is formed of which EF and EG are sides; but in order to follow the same argument as in Case II. we must produce the equal sides DF, DG to K and H, then the angles on the other side of the base of the isosceles triangle DFG become the medium of comparison, and the rest of the proof is identical with that of Case II. D F E G H K be This case may very simply proved without any construction by the help of Prop. 21; for since DF, FE are drawn 'from the ends of one side of a triangle to a point within it,' they are less than DG, GE, but DF = DG, therefore EF is less than EG. 6 Euclid himself only notices the second of these cases. In order to remedy this imperfection, Dr. Simson in his edition reduces the three cases to one, by inserting into the text the words, of the two sides DE, DF, let DG be the side which is not greater than the other.' This condition ensures EF falling below EG, for, says Dr. Simson, 'it is easy to perceive that DG being equal to DF, the point G is in the circumference of a circle described from the centre D at the distance DF, and must be in that part of it which is above the straight line EF, since the angle EDG is greater than the angle EDF.' We will now pass to the summary of this proposition, taking the three cases, which seems to be on the whole the most satisfactory method. If you wish to follow Dr. Simson, you must remember to make the condition above referred to, that DE is not greater than DF. SUMMARY OF PROPOSITION XXIV., THEOREM 15. If two triangles have two sides of the one equal to two sides of the other, but the angle contained by the two sides of the one greater than the angle contained by the two sides equal to them of the other, the base of that which has the greater angle must be greater than the base of the other. Cons.-At point D in a DEF make EDG = < BAC, making DG = AC (I. 23). Join EG, GF. Case III. produce DF and DG to H and K. Proof.-A DEG = ▲ ABC (I. 4). Case I. When EF falls on EG; EF is a part of EG.. EG is greater than EF. Case II. When EF falls below EG; .. EG is greater than EF (I. 19), but EG = BC Case III. GFK = ▲ FGH, and proceed as in Case II. I. 25. The converse proposition follows, which is demonstrated, as is usually the case, indirectly. The triangle with the larger base has the larger vertical angle, if not either I. 4 or the last proposition is contradicted. SUMMARY OF PROPOSITION XXV., THEOREM 16. If two triangles have two sides of the one equal to two sides of the other, each to each, but the base of the one greater than the base of the other, the angle contained by the sides of that which has the greater base shall be greater than the angle contained by the sides equal to them of the other. Proposition XXVI. demonstrates the third case of the equality of triangles considered by Euclid. It shows that triangles are equal, if in each triangle two angles and one side are equal to the corresponding angles and side in the others. The proof, as in the other propositions treating of the equality of triangles, is indirect. Let us take two triangles, ABC, DEF, of which the angles ABC, BCA, are severally equal to the angles DEF, EFD. You will observe that the D F equal sides may be related to the equal angles in two ways: they may lie between them as the sides BC and EF, or they may be opposite to one of the equal angles, as either of the other These relations of sides and angles give two cases in the theorem requiring a separate proof. I. Let us take the case in which the equal sides lie between (or 'are adjacent' to) the equal angles. We have, as you will remember, two means of showing the equality of triangles, i.e. by satisfying the conditions of I. 4, or of I. 8. A D If Since the angles at B and C are equal to those at E and F, and BC = EF, if we can show that either of the other pairs of sides are equal, we shall have the conditions required by I. 4, and it follows that the triangles in question are equal. Let us then endeavour to show that AB = DE, or, as we are to follow the indirect method, that some absurdity will result if these lines are not equal. AB is not equal to DE, one of them must be greater than the other; let AB be the greater, and from it cut off a part GB equal to DE. Then by joining GC it is plain that a triangle in all respects to DEF (I. 4), for we EF, and the included angles were E F GBC is formed equal have GB, BC= DE, given equal. Since the triangles GBC and DEF are equal in all respects, the angle GCB = angle DFE, but we know also that |