228°. ABC is a triangle in its circumcircle whose diameter we will denote by d. Hence the sides of a triangle are proportional to the sines of the opposite angles; and the diameter of the circumcircle is the quotient arising from dividing any side by the sine of the angle opposite that side. PRINCIPLE OF ORTHOGONAL PROJECTION. 229°. The orthogonal projection (167°, 2) of PQ on L is P'Q', the segment intercepted between the feet of the perpendiculars PP' and Q2'. P P' L Now P'Q' PQ cos (PQ. P'Q'). .. the projection of any segment on a given line is the segment multiplied by the cosine of the angle which it makes with the given line. From left to right being considered as the + direction along L, the segment PQ lies in the 1st Q., as may readily be seen by considering P, the point from which we read the segment, as being the centre of a circle through Q. Similarly QP lies in the 3rd Q., and hence the projection of PQ on L is + while that of QP is -. When PQ is to L, its projection on L is zero, and when to L this projection is PQ itself. Results obtained through orthogonal projection are universally true for all angles, but the greatest care must be exercised with regard to the signs of angular functions concerned. Ex. AX and OY are fixed lines at right angles, and AQ is any line and P any point. Required to find the LPQ in terms of AX, PX, and the LA. X Take PQ as the positive direction, and project the closed figure PQAXP on the line of PQ. Then pr.PQ+pr.QA+pr. AX+pr.XP=0. (168°) PQ, and pr.QA=0; AX lies in 1st Q., and P N Now, pr.PQ is XP in the 2nd Q. Moreover (AX. PQ), i.e., the rotation which brings AX to PQ in direction is .'. LN, and its cosine is +. cos (AX. PQ) = +sin A. Also, pr.XP is - XP cos LXPQ= − XP cos A. PQ=XP cos A - AX sin A. SIGNS OF THE SEGMENTS OF DIVIDED 230°. AOB is a given angle and LAOB= −▲BOA. Let OP divide the LAOB internally, and OQ divide it ex ternally into parts denoted respectively by a, ẞ, and a'ß'. B = LAOQ, and If a is the LAOP and ẞ the 4POB, a and are both positive. But if we write Q for P, a' B'=LQOB, and a' and ' have contrary signs. On the other hand, if a is LAOP and ẞ the BOP, a and ẞ have contrary signs, while replacing P by Q gives a' and p' with like signs. The choice between these usages must depend upon convenience; and as it is more symmetrical with a two-letter notation to write AOP, BOP, AOQ, BOQ, than AOP, POB, etc., we adopt the convention that internal division of an angle gives segments with opposite signs, while external division gives segments with like signs. In like manner the internal division of the segment AB gives parts AP, BP having unlike signs, while external division gives parts AQ, BQ having like signs. Def.-A set of points on a line is called a range, and the line is called its axis. By connecting the points of the range with any point not on its axis we obtain a corresponding pencil. (203°, Def.) Cor. To any range corresponds a pencil for every vertex, and to any vertex corresponds a range for every axis, the axis being a transversal to the rays of the pencil, If the vertex is on the axis the rays are coincident; and if the axis passes through the vertex the points are coincident. 231°. BY is any line dividing the angle B, and CR, AP are perpendiculars upon BY. B Then ΔΑΡΥ= ΔCRY, and and .. Thence Therefore a line through the vertex of a triangle divides the base into segments which are proportional to the products of each conterminous side multiplied by the sine of the corresponding segment of the vertical angle. Cor. 1. Let BY bisect LB, then AY YC = AY={(6—AY), and AY= a YC= с a R bc a + c ba a + c Y P Which are the segments into which the bisector of the B divides the base AC. Cor. 2. In the ▲ABY, BY2-AB2+AY2-2AB. AY.cos A. b2+c2- a2 bc 2bc a+c But cos A whence by reduction BY2= (217°), and AY= = b ac{1-(~24-)"}, I ·a+c which is the square of the length of the bisector. =αc Cor. 3. When AY=CY, BY is a median, and AB sin YBC .. a median to a triangle divides the angle through which it passes into parts whose sines are reciprocally as the conterminous sides. 232°. In any range, when we consider both sign and magnitude, the sum AB+BC+CD+DE+EA=0, however the points may be arranged. (217°) For, since we start from A and return to A, the translation in a direction must be equal to that in a direction. That this holds for any number of points is readily seen. Also, in any pencil, when we consider both sign and magnitude, the sum LAOB+<BOC+4COD+4DOA=0. For we start from the ray OA and end with the ray OA, and hence the rotation in a + direction is equal to that in a - direction. RANGES AND PENCILS OF FOUR. 233°. Let A, B, C, P be a range of four, then AB. CP+BC. AP+CA. BP=0. AP=AC+CP, and BP = BC+CP. Proof... the expression becomes BC(AC+CA)+(AB+BC+CA)CP, and each of the brackets is zero (232°). .. etc. 234°. Let O. ABCP be a pencil of four. Then sin AOB. sin COP+ sin BOC. sin AOP+sin COA.sin BOP=o. Proof.—▲AOB=OA. OB sin AOB, AAOB AB. p, is the common altitude to all also where AB.p=OA. OB. sin AOB. Similarly, CP.p=OC.OP.sin COP. A B A'P B AB.CP.p2=OA. OB. OC. OP sin AOB. sin COP. Now, p2 and OA. OB. OC. OP appear in every homologous product, ... (AB. CP+BC. AP+CA.BP)p2 B'P P =OA.OB.OC. OP(sin AOB. sin COP + sin BOC. sin AOP + sin COA. sin BOP). But the bracket on the left is zero (233°), and OA.OB.OC.OP is not zero, therefore the bracket on the right is zero. q.e.d. 235°. From P let perpendiculars PA', PB', PC' be drawn to OA, OB, and OC respectively. Then sin AOP. sin BOP=. etc., and putting these values for sin AOP, etc., in the relation of 234°, we have, after multiplying through by OP, C'P. sin AOB+A'P. sin BOC+ B'P. sin COA = Or, let L, M, and N be any three concurrent lines, l, m, n the perpendiculars from any point P upon L, M, and N respectively, then /sin MN+m sin NL+n sin LM = 0. where MN denotes the angle between M and N, etc. |