Construction. Besides joining the vertices C, D, produce the two sides A C and A D to E and F. E B Demonstration. 1. Then because AC is equal to AD, the angle ECD is equal to the angle F D C (Prop. 5). 2. But the angle ECD is greater than the angle BCD (Ax. 9). 3. Therefore also the angle FDC is greater than the angle B CD. 4. Much more is the angle BDC greater than the angle B C D. 5. On the other hand, because BC is equal to BD, the angle BDC is equal to the angle BCD (Prop. 5). 6. That is, if the vertex of one of the triangles falls within the other, the same angles B CD, B DC, are equal and unequal at the same time, which is impossible; therefore the vertex of one of the triangles does not fall within the other. The case in which the vertex D of one of the triangles ADB is on the side of the other triangle A CB, needs no demonstration, for it is evident from Ax. 9 that BC cannot A be equal to B D. Therefore on the same base and on the same side of it there cannot be two triangles which have the two sides terminated in one extremity of the base equal to each other, and likewise the two sides terminated in the other extremity. Q.E.D. From the Seventh Proposition, Euclid immediately deduces the Eighth. After the foregoing discussion and explanation, it will be sufficient to write it out formally, without any further remarks. THE EIGHTH PROPOSITION WRITTEN OUT. PROPOSITION VIII. THEOREM. General Enunciation. It is given that two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal: It is required to prove that the angle contained by the two sides of one of the triangles is equal to the angle contained by the two sides equal to them of the other. Particular Enunciation. Let ABC and DEF be two triangles which have the two sides BA, AC of the one equal to the two sides E D, D F of the other, each to each, viz. BA to E D, and AC to DF; and have likewise the base B C equal to the base EF: It is required to prove that the included angle BAC is equal to the included angle E D F. Construction. Let the triangle A B C be applied to the triangle DEF, so that the point B may be on E, and the base B C may lie along E F. Demonstration. Then shall the point C come on the point F, because BC is equal to E F; and the bases BC, EF will occupy the same space. This being the case, the two sides B A, A C must lie along the two sides E D, D F, for if not they will have a different situation as E G, G F. And then, on the same base and on the same side of it, there will be two triangles which have the two sides terminated in one extremity of the base equal to each other, and likewise the two sides terminated in the other extremity. But this is impossible (Prop. 7). Therefore the two sides BA, A C cannot but lie on the two sides ED, DF, and therefore the angles BAC and EDF fill the same space, and are equal (Ax. 8). Wherefore, if two triangles have two sides of the one equal to two sides of the other, each to each, and have likewise their bases equal; then shall the angle contained by the two sides of the one be equal to the angle contained by the two sides equal to them of the other. Q.E.D. Euclid might have proceeded to show that the two remaining angles are also equal, as well as the included areas, for he has here shown, as completely as he did in the Fourth Proposition, that the two remaining angles and the included areas respectively fill the same spaces. He does not, however, proceed any further, nor is it necessary, for when he has proved that the included angles are equal, he has before him two triangles which have two sides of the one equal to two sides of the other, and likewise the included angles equal: and therefore, by the Fourth Proposition, the two triangles are equal in every respect. The learner, having proved the Eighth Proposition as well as the Fourth, can now go forward with these two propositions, one in either hand, engines, as it were, of use and power in the solution of other problems and theorems. The next four propositions are problems, to be solved by one or other of these two propositions. But the learner must be careful to apply them rightly. If it is the included angles which have to be proved equal, he must apply the Eighth Proposition. If it is the bases which have to be proved equal, he must apply the Fourth Proposition. In both cases the first three of the five steps of reasoning, often repeated in the exercises on the Fourth Proposition, are the same, whether it is the Fourth or the Eighth Proposition which is applied. The last two steps change places. In applying the Fourth Proposition, the fourth and the fifth steps of the reasoning are, (4) And the included angles [naming them] are equal [the reason why these angles are known to be equal being added]. And the conclusion is: (5) Therefore by the Fourth Proposition the bases are equal. Q.E.D. In applying the Eighth Proposition, the last two steps of the reasoning are as follows: (4) And the bases are equal [the reason why they are known to be equal being added]. And the conclusion is: (5) Therefore, by the Eighth Proposition, the included angles are equal. Q.E.D. THE NINTH PROPOSITION DISCUSSED. With these preliminary remarks we turn to the Ninth Proposition, the enunciation of which is as follows: Given a rectilineal angle :— It is required to bisect it. Euclid draws an angle and calls it the angle BAC, and adds, It is required to bisect B A C. His construction is as follows: And by the Third Proposition be cuts off from AC a part A E equal to A D. He joins D E. And on the side of DE, remote from A, he describes an equilateral triangle D F E by Prop. I. [You will be told by and by why he says on the side of DE remote from A.] He adds, Join A F. And he concludes thus: Then will the angle B A C be bisected by the straight line A F. Before exhibiting to the learner the demonstration of this proposition in a formal manner, the following suggestions are given, to encourage him to make out the demonstration for himself. |