less than the sum of the sides AC and CB, and that the APB is greater than the ZACB. To prove this, produce the line AP, to meet the side CB in the point D. 1. In the ▲ ACD, the sum of the sides AC and CD is greater A than the side A D. C D B Let the line D B be added to each of these unequals, —that is, to the sum of A C and CD, and to A D. It will follow that the sum of A C, CD, and DB, will be greater than the sum of A D and D B. But the sum of A C, CD, and D B is the same as the sum of A C and CB. Therefore, the sum of the lines A C and C B is greater than the sum of the lines A D and DB. Again, in the ▲ PDB, the sum of the sides BD and D P is greater than the side B P. Let the line PA be added to each of these unequals, —that is, to the sum of B D and D P and to BP. It will follow that the sum of BD, DP, and P A, will be greater than the sum of P B and P A. But the sum of BD, DP, and PA, is the same as the sum of BD and D A. Therefore, the sum of BD and DA is greater than the sum of BP and PA. Now it was before shown that the sum of A C and CB is greater than the sum of A D and D B. Much more, therefore, is the sum of A C and CB greater than the sum of A P and P B, which is less than the sum of AD and D B. 2. On looking at the ▲ B D P, with its produced side D P A, it will be seen that the APB is an exterior of that A, and the PDB one of the interior opposites. Consequently (Prop. XVI.), the APB is greater than the PD B. · Again, the PDB is an exterior of the ▲ ACD, ACD. formed by the production of the side C D. The PDB, therefore, is greater than the But it was before shown that the APB is greater than the PD B. Much more, therefore, must the ZAP B be greater than the AC B, which is less than the PD B. In proving this proposition, it would have done equally well if the line BP had been produced to meet the side A C in E. E B It would then have been shown that the sum of AC and CB is greater than that of AE and EB; and the sum of AE and EB greater than that of AP and P B. From which it would have been inferred that the sum of AC and CB is greater than that of AP and P B. Secondly, it would have been shown that the exterior APB is greater than the interior opposite AEP; and that the AE P is, in its turn, an exterior of the ▲ ECB, and, therefore, greater than the interior and opposite, ECB. From which it would have been inferred, that the APB is greater than the ACB. PROPOSITION XXII. To make a triangle, the sides of which shall be equal respectively to three given straight lines, any two of which are greater than the third. For the construction in this proposition we must be able, 1. To describe a circle with a given centre, at a given distance from that centre. (Post. III.) 2. To join two given points by a straight line. (Post I.) 3. From a given point to draw a straight line equal in length to a given finite straight line. (I. 2.) 4. From a given line to cut off a part equal in length to a given finite right line. (Prop. III.) To prove that the construction effects what is required we must know, 1. That all radii of the same circle are equal in length. (Def. XV.) 2. That magnitudes which are equal to the same are equal to one another. (Ax. I.) Let AB, CD, and EF be the three given lines any two of which are greater than the third. KM, equal to AB. (Prop. III.) From the point K draw a line equal in length to the line CD. Let it be the line KL. (Prop. II.) From the point M draw a line, MN, equal in length to the line EF. With K as centre, and at the distance KL, describe a circle. (Post. III.) With M as centre, and at the distance M N, describe another circle. These two circles cannot lie outside each other, but must be partly within and partly without each other. For (P being the point where the circle LPO intersects the line GH) KP is equal to KL, and is there F fore equal to CD. Now MP must be less than EF, and therefore less than a radius of the circle described with the centre M, because CD and EF together are greater than AB, and therefore greater than K M. Consequently, KP and a radius of the circle described with the centre M must be greater than KM; i.e., MP is less than a radius of the circle described with the contre M. It follows, therefore, that a part of each circle lies within the other. Consequently, they cut each other in two points. (See Ax. C.) Take O, one of these points of intersection, and join K and O by the right line KO, and M and O by the right line MO. The AKOM will be the ▲ required. For the side KO is equal to the line KL, both being radii of the same circle. But KL was drawn equal to CD. Therefore also KO is equal to CD. Again, the side M O is equal to MN, both being radii of the same circle. But MN was drawn equal to EF. Therefore, MO is equal to E F. Lastly, the side KM was cut off of the same length as AB. It has thus been shown, that the three sides of the AKOM are equal respectively to the three given straight lines A B, CD, and EF. PROPOSITION XXIII. From a given point in a straight line to draw a straight line making with the given straight line an angle equal to a given angle. For the construction employed in this proposition we must be able,— 1. To join two points by a straight line. (Post. I.) 1. That radii of the same circle are equal. (Def. 2. That magnitudes which are equal to the same are equal to one another. (Ax. I.) 3. That two sides of a ▲ are together greater than the third. (Prop. XX.) 4. That if two As have the three sides of the one equal respectively to the three sides of the other, the As will also be equal in every other respect. (Prop. VIII.) Let B A C be the given rectilineal Let DE be the given straight line, and D the given point in it. It is required to draw from the point D a straight line, which shall make with the line DH an equal to the < BAC. In the line A C take any point, C. Join the points C and B by the straight line C B. From the point D draw a line, DF, equal in length to the line A C. (Prop. II.) From the line DE cut off a part, D H, equal in length to the line A B. (Prop. III.) |