Join PQ. APQ will then be the required triangle. The result is evident, but should be shown by the pupil. PROBLEM IX. 186. Two angles of a triangle being given, to con struct the third angle. Given. Two angles, c and e, of a triangle. Required. To find the third angle of the triangle. Construction. 1. At any point O in an indefinite line AB make the angle BOC equal to the given angle c. 2. At the same point make the angle COD gle e. given an A D e Then the angle DOA will be the angle required. B Proof. From Theorem IV., to be supplied by the student. PROBLEM X. 187. To construct a triangle, having given one side and the two adjacent angles. 2. At A make the angle BAM = angle c. 3. At B make the angle ABN = angle e. B e 4. Continue the lines AM and BN until they meet, and let C be their point of meeting. ABC will then be the required triangle. The proof should be supplied by the student. Corollary. Since the third angle of a triangle can always be found when two angles are given, this problem, combined with the preceding one, will suffice for the construction of the triangle when one side and any two angles are given. PROBLEM XI. 188. Two sides of a triangle and the angle opposite one of them being given, to construct the triangle. Given. The two sides, a BR cut off a length BC equal to the given line a, which is not opposite the given angle B. 3. From C as a centre, with a radius equal to the line b describe a circle cutting BM at the points D and D'. Either of the triangles BCD or BCD' will then be the required triangle. The proof follows at once from the construction. 189. Scholium. The fact that there may be two triangles formed from the given data has been explained in the scholium § 113. PROBLEM XII. 190. Through a given point to draw a straight line which shall be parallel to a given straight line. Given. A straight line, AB; a point, P. Required. To draw a straight line through P parallel to AB. Construction. 1. Take any point D in AB, and join PD. Proof. By Theorem III., because the angles PED and PDB are alternate angles. PROBLEM XIII. 191. To divide a finite straight line into any given number of equal parts. Given. A finite straight line, AB; a number, n. Required. To divide AB into n equal parts. Construction. 1. From one end of AB draw an indefinite straight line, making an angle with AB different from a straight A angle. n 2. Upon the indefinite line lay off any equal lengths, A 1, 12, 23, etc., until n lengths are laid off. 3. Join B to the end n of the last length. 4. Through each of the points 1, 2, 3, parallel to nB intersecting AB. n, draw a The line AB will then be divided into n equal parts by the points of intersection. Proof. The parallels intercept equal lengths on the transversal An, by construction. Therefore they also intercept equal lengths on AB, and the number of intercepted lengths is n (§ 129). Therefore AB is divided into n equal parts. Q.E.F. PROBLEM XIV. 192. Two adjacent sides of a parallelogram and the angle which they contain being given, to describe the parallelogram. Given. Two lines, AC and GH; an angle, O. Required. To form a parallelogram having GH and AC for two adjacent sides, and O for the angle between these sides. Construction. 1. At one end A A of the line AC make an angle o equal to 0. -H 2. On the side of this angle take AB = the given line GH. 3. Through B draw a line BD parallel to AC. 4. Through C draw CD parallel to AB, intersecting BD in D. ABCD will be the required parallelogram. Proof. May be supplied by the student. 193. Corollary. To construct a square upon a given straight line. This problem is a special case of the preceding one, in which the given sides are equal and the given angle is a right angle. To solve it: Draw AC and BD 1 AB. And the square will be complete. CHAPTER VI. EXERCISES IN DEMONSTRATING THEOREMS. The following theorems should be demonstrated by the student in his own way, so far as he is able. Analysis of a given Theorem. The first step in the process of finding a demonstration is to state the hypothesis, referring to a diagram, which it is generally best the pupil should draw for himself. The statement should include not simply what the theorem says, but what it implies. Reference must be made to definitions until the hypothesis is resolved into its first elements. Next, the conclusion must be analyzed in the same way, in order to see not only what it says, but also what it implies. By the analyses of these two statements they must as it were be brought together, in order to see in what way they are related. The process of discovering this relation is one which the student must find for himself in each case, and for which no rule can be given. Frequently, however, it will be necessary to draw additional lines in the figure, and to call to mind the various theorems which apply to the figures thus formed. To facilitate this, references to previous theorems which come into play are added. The relation being found, the demonstration must next be constructed in the simplest manner, but without the omission of any logical step. This, also, is a matter of practice in which no general rule can be given. It is recommended that the teacher require the pupil to express each step of the demonstration with entire completeness and fullness. Some of the first theorems are so simple that the only serious exercise is that of constructing an artistic demonstration. The work thus becomes a valuable exercise in language and expression as well as in geometry. The most common fault is that of passing over steps in the demonstration because the conclusion seems to be obvious. One of the great objects of practice in geometry is to cultivate the habit of examining the logical foundations of those conclusions which are accepted without |