А B same side of it, then if the two sides A C and A I), terminating at A, were equal to each other, the sides BC and B D, terminating at the C D other extremity of the base, could not also be equal to each other; or if BC and B D were equal, then AO and A D could not also be equal to each other. It is not meant that neither pair of conterminous sides can be equal, but that both pairs cannot be equal at the same time.* This proposition is proved by showing that if we were to suppose it possible that there could be two such As as are described in the enunciation, we should be compelled to admit something to be true which is obviously impossible. For the proof of the proposition we require,- given point to another. (Post. I.) length. (Post. II.) We must also know, 1. That if a magnitude be less than one of two equal magnitudes, it is also less than the other. 2. That if a magnitudo be less than the smaller of two unequal magnitudes, it is also less than the larger. 3. That if two sides of a A are equal, the s The beginner must not suppose that an elaborate proof of this proposition is unnecessary, because the inequality of one or other pair of sides is obvious to the eye. Geometrical truths are never to be settled by this test. And, in fact, if a scale sufficiently large were adopted, it would be possible to draw two As placed like those in the diagram, such that the inequality of the sides could scarcely be detected by the most accurate measurement. D B opposite to them are also equal, and if the equal sides be produced, the <s on the other side of the base are likewise equal (Prop. V.). C D Let us suppose that there could be two As A C B and A D B, on the same base A B, and on the same side of it, and such that A C and A D are equal to one another, and also BC and B D equal to one another. * First let the As be such that the vertex of each falls outside the other. Join the points C and D by the right line C D. In the A AC D, if AO and A D be equal to one another, the < AU D is equal to the < A DO (Prop. V.) If the sides B C and B D could also be equal to one another, then also in the A B C D the < B C D would be equal to the < B D C. But the < B C D is less than the < ACD (being a part of it). Therefore the < B C D must also be less than the < ADC, which is equal to the < ACD. But the < ADC is less than the < B D C (being a part of it). Therefore the < B C D, which is less than the 2 A D C, must also be less than the 2 BD C. But it was shown before, that if the sides B C and BD be equal, the 8 B C D and B D C are equal to each other. Thus the supposition that, in the As A B C and A B D, the sides, A C and A D, could be equal to * The beginner must beware of supposing that the cbject of the proposition is to prove that the sides A C and AD are not equal to the sides B C and BD; a mistake which the writer has found made more frequently than might be supposod. each other, and that the two sides B C and B D could also be equal to each other, renders it necessary that we should admit that two Zs (namely B C D and BD C) are both equal and unequal at the same time, —which is an impossibility. Now the supposition which leads us of necessity to this absurdity, must itself be absurd. That is, it is absurd to suppose that the two pairs of sides, A C and A D, and B C and B D, can be equal at the same time. (If one pair be equal, the other pair must be unequal.) Next let us suppose it possible that there should be two Aš on the same base and on the same side of it, having the sides that terminate at one extremity of the base equal, and likewise those that terminate at the other extremity of the base equal, so placed that the vertex of one of the As (A B C) falls within the other A (A B D). Join the points D and C by the right line D O. Produce the line A D to any point E, and the line A C to any point F. If it were possible that the sides A D and AC should be equal to one another, and also the sides B C and B D equal to one another, then the ADAC would be an isosceles A, and therefore the <8 E D C and F C D, formed by producing the equal sides A D and A C, would be equal to one another; and, in the ACB D, the s B C D and B D C would be equal to one another. But the < B D C is less than the < EDC (being a part of it): therefore the < B D C is also less than the _ FCD, which is equal to E D C. Much more then is the < B D C less than the < B C D, which is greater than the <FCD. F E D B A But it was before shown that, if we suppose the sides B C and B D to be equal to each other, as well as A D and A C, then the < BDC must be equal to the < BCD. But it is impossible that the same Z8 (B D C and BCD) should be both equal and unequal. Consequently, the supposition which led to this impossibility must be absurd. That is, it is absurd to suppose that, on the same base, and on the same side of it, there can be two As, having the sides terminating in one extremity of the base equal to one another, and likewise those terminating in the other extremity of the base equal to one another. PROPOSITION VIII. If two triangles have the three sides of the one equal to the three sides of the other, each to each, then the triangles will also be equal to each other in every other respect, that is, the angles of the triangles will be equal, each to each, and the triangles will be equal to each other in area. For proving this proposition we must know,- upon the other as to coincide in direction, and for any portion of their length. (Ax. B.) 2. That upon the same base, and upon the same side of it, there cannot be two as having both pairs of conterminous sides equal. (Prop. VII.) Let A B C and D E F be two As, having the side A B equal to the side D Е, the side A C to the side D F, and the side C B to the side F E. B D A B The proof of the proposition consists in showing that the two As may be placed one on the other, so as to coincide in every part. Place the point D on the point B, and let the line D E lie along the line A B. Then, because the line D E is equal to the line A B, the point E will fall upon the point B.* Let the As lie on the same side of the base A B. Then the line D F must lie along the line A C, and the line E F along BC, the point F coinciding with the point C. For if they did not, but lay in any other position, as, for example, in the CF annexed figure, we should have two As, A C B and A F B, standing on the same base, and on the same side of it, and having the conterminous sides A C and A F equal, and likewise the conterminous sides B C and B F equal. This was proved in the last proposition to be impossible. Consequently, the supposition which leads to this impossibility must itself be impossible. That is, it is impossible, when the side D E has been placed so as to coincide with the side A B, that the side D F should lie in any other position than along A C, and the side E F in any other position than along B C. Therefore, as they must lie in some position or other, DF must lie along A C, and E F along B C. Consequently, as the sides F D and D E coincide in direction with C A and A B, the < FDE is equal to the < C A B. In like manner the < FED is equal to the ZOBA, and the < D F E to the < ACB, and the area of the one A is equal to the area of the other. Beware of saying, “Place the point D upon the point A, and let the line D E lie along the line A B and let the print E fall on the point B.” (See p. 9.) |