From the point H draw a line, HG, equal in length . to the line BC. (Prop. II.) With D as centre and at the distance D F describe a circle. (Post. III.) With H as centre and HG as radius describe another circle. Because AC and CB are together greater than A B (any two sides of a ▲ being greater than the third side), and AC, CB, and AB are equal respectively to DF, HG, and D H, it follows that D F and H G are together greater than D H. Consequently the two circles described with the centres D and H will intersect one another. Let K be one of the points where they intersect. DK will be the line required. For, join K and H by the right line KH. In the ▲ DKH the side DK is equal to the line DF, both being radii of the same circle. But D F is equal to A C. Therefore DK is equal to A C. (Ax. I.) But HG is equal to CB. Therefore KH is equal to CB. And lastly, the side DH was cut off of the same length as A B. Thus, in the As BAC and HDK, the sides of the one are equal respectively to the sides of the other. Consequently the As are also equal in every other respect. Among these respects is that the KDH is equal to the CA B. That is to say, the line D K has been drawn so as to make with the line DE an equal to the given CAB. The directions for the construction, as set down in the ordinary editions of Euclid, involve making a triangle, the three sides of which shall be equal respectively to the three sides of another triangle, and, moreover, so that the sides shall lie in a given position. Now the 22nd proposition does not provide for this, and the construction as given above simply states explicitly what is necessary to effect this. It will be convenient, before entering upon the 24th proposition, to establish the following introductory proposition, of which use is made in the course of the 24th proposition. If a straight line be drawn from the vertex of a triangle to any point in the base between its extremities, that line will be shorter than one or other of the other two sides of the triangle.* This proposition admits of two cases :— 1. Where the ▲ is an isosceles A. 2. Where one of the two sides is longer than the other. 1. Let ACB be an isosceles A, and CD a line drawn from the vertex to the point D in the base It has to be shown that the line CD is less than either of the sides A C, CB. The CDA is the exterior of the A CBD, formed by the production of A D the side BD. C Therefore, the CDA is greater than the CBD. B * It may be less than either-but must be less than one or other. But since A CB is an isosceles A, the CAB is equal to the CBA. (Prop. V.) Consequently the CDA, which is greater than the CBA, is also greater than the CAB. But since, in the ▲ CD A, the CDA is greater than the CAD, it follows (according to the 19th proposition) that the side CA is longer than the side CD. And since CB is equal to CA, CB is also longer than CD. 2. Let ACB be a ▲ in which the side AC is longer than the side CB. It has to be shown that the line C D is shorter than the longer of the two sides. Because CA is longer than C B, it follows (according to the 18th proposition) that the CBA is greater thethan C A B. CDA is the exterior of the A CBD formed by the production of Consequently the CD A is greater than the CBD. Much more therefore is the CDA greater than the CAD, which is less than the CBD. But since, in the ▲ CD A, the CDA is greater than the CAD, it follows that the side CA is longer than the side CD. In each of these two cases it will be observed, that the side with which the line drawn to the base is compared may be described as that side which is not the less of the two. And what has been demonstrated in the two cases may be summed up by saying, that "If a line be drawn from the vertex of a triangle to any point of the base between its extremities, that line is shorter than that one of the other two sides of the triangle which is not the less of the two." PROPOSITION XXIV. 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 said two sides of one of them greater than the angle contained by the said two sides of the other, the base of that triangle which has the greater angle will be greater than the base of the other. For the construction necessary in this proposition 1. To join two given points by a straight line. 3. From a given point in a given straight line to For proving the proposition by the aid of the construction we must know, 1. That if a line be drawn from the vertex of a ▲ to a point in the base, that line is less than that 2. That if two As have two sides of the one equal Let ABC and DEF be two As, in which the sides AB and BC are equal to the sides DE and which is not the greater of the two (i.e., if the As are not isosceles, let it be the shorter of the two sides). From the point B draw a line, B L, making with the line A B an equal to the DEF, and meeting the base AC in the point L. BL being a line drawn from the vertex of a ▲ to a point in the base between its extremities, is shorter than that one of the two sides of the ▲ ABC which is not the less of the two. That is to say, it is shorter than BC (which we suppose to be either equal to A B, or longer than A B). Produce BL to any length, and from BL produced cut off a part, BH, equal in length to B C.* B H * It is essential for the proof of the proposition that the point H should be on the opposite side of the base A C to the vertex B. For this purpose it is necessary that the line drawn from B, making with one of the sides of the ▲ A B C an equal to the DEF, should make that with that one of the two sides, AB, BC, which is not the longer of the two. If BC be longer than BA, and the line BH be drawn making with BC an equal to the DEF, it is quite possible that the extremity, H, of the line cut off equal to BA may fall within the ▲; in which case the construction does not enable us to establish the proposition. C |