Page:Encyclopædia Britannica, Ninth Edition, v. 10.djvu/408

Rh 394 G E O M to a given r.i_v c in S1, is found by joining S, to the point where c ' cuts the axis s. A similar construction holds in the other eascsof pcrspectivctiguies. (hi this depends the solution of the following general problem. §36. ProbIcIn.—Tliree pairs of corresponding elements in two projective rows or pencils be.iiig given, to determine for any element in ope the corresponding element in the other. e solve this in the two cases of two projective rows and of two projective flat pencils in a plane. {’:;oblem. I.-—I.ct A13“. C ‘he three pom 9 in a row 8, .- '. '. t" t ie corre- sponding points in a projective row s‘, both being in a plane; it is re- quired to find _for any point 1) in s is required to ﬁnd for any ray (1 in S the corresponding point D’ in s’. the corrcsponiling ray cl’ in S’. The solution is made to depend on the construction of an auxili- ary row or pencil which is perspective to both the given ones. This is found as follows :— Solution. 0 Problem I.——On the line 'oinino' two corres iondinrr _ I _ .l _ o M h points, say AA (fig. 10), take any two points, S and S, as centres of auxiliary pencils. Join the iiitersec- tion B, of SB aml S'B’ to the int_erscc- tion (‘I of SC and II I ' .30 by the line sl. Then a row on 31 will he perspective to swith S as centre of projection, and to s’ with S’ as centre. To ﬁnd now the point D’ on s’ corre- sponding to a point e D on s we have only “ ‘ to determine the point D1, where the line SD cuts s1, and ‘ to _draw S’D, _; the point vhere tliisliiie cuts 3' will be the required point D’. Proof. —The rows 3 and s’ are both . perspective to the E3‘ 10' row sl, hence they are projcct.ivc to one another. To A, B, C, D on 5' correspond A1, B‘, C,, D1 on s,, and to these correspond A’, 13’, C’, D’ on s’; so that D and D’ are corresponding points as required. Solution of Problem II.—Tlirougli the intersection A of two cor- responding rays a and a’ (ﬁg. 11), take two lines, s and s’, as bases of auxiliary rows. Let S, be the point where the line b,, which joins I} and _l3’, cuts the lnic c], which joins C'1t£.llﬂ C’. Thgn at pc§cil_.tS11 wi e perspcc ive o vi 1 s as axis of projection . To find the ray cl’ in S’ corresponding to a given ray d in S, cut cl by sat D ; project this ponit from S, to D_’ on s’ and join D’ to S’. This will he the required ray. _ I’/'aQf._—:_'l‘lia1t tshe ]&('lli‘ll S, "r°.§i’1lI‘(.*s’}§7’o..‘i ..‘.‘..'.‘.t.-1”}.-T;’...’.‘.’ To the lines a,, b_,, cl, d, in S, corgespolntl} thle llllC§(1[,/b,’C, all, in. am ie mcs a,, c ¢ in S’, so‘ that cl and cl’ ’are corresponding rays. In the first solution the two centres, S, S’, are anytwopoints Problem II.—Let a-.b.c be three rays in a pencil S, a’, b’. c’ the corre- spondin.-.1: rays in a projective pencil S’, both being in the same plane; it 6‘: 4 — _— Ar

Fig. 11. on a line joining any two corresponding points, so that the solution of the problem allows of a great many different constructions. But whatever construction be used, the ;ooi7i.t I)’, cor7'cspo11d2'ng to 1), 1n'u..s-t be always the same, according to the theorem in § 29. This gives rise to a number of thcorcins, into which, however, we shall not enter. The same remarks hold for the second problem. § 37. As a further application of the theorems about perspective rows and pencils we shall prove the following important theorem. Theorem.—l f A IEC and A'B’U’ (fig. 12) be two triangles, such that the lines AA’, 3 3’, CC’ meet in a point S, then the intersections of BC and B'C’, of CA and C’A’, and of AB and A’B’ will lie in a line. Proq/'.—Let a, b, c denote the lines AA’, BB’, CC’, which meet at S. Th--n these may be taken as hases of projective rows, so that A, A’, S on a correspond to B, B’, S on b, and to C‘, C’, S on c. As the point S is common to all, any two of these rows will be perspective. E T R Y [cnoJEcrivE. If S, be the centre of projection of rows b and e, S‘-_» ,, ,. ,, c and a, >3 ,, ,, ,, u and I), and if the line SISE cuts (L in Al, and b in l‘:,, and c in C1, then A,, B, will be corresponding points in a and In, both corresponding to C1 in c. llut (C aml b are perspective, thercl'orc the line Allll, that is S,S:, joining correspond- ing points must pass through the centre of pro- jection S3 of a and b. In other words, S1, S3, S3 lie in a line. This is I)csargues’s celebrated theorem if we state it tlius:—— Theorem of Dcsar_r/ucs. ~lf each of two triangles has one vertex on each of three concurrent lincs, then the intersections of corresponding sides lie in a line, those sides hcing called correspond- ing which are opposite to vcrticcs on the same. line. The converse theorcm holds also, viz.:— T/zrurcm.——If' thc sides of one triangle meet those of another in three points which lie in a line, then the vcrticcs lie on three lines which meet in a point. The proof is almost the same as before. § 38. .l[«'t)'z'c(Ll rclrztioizs belzcccn ]my'cclii'e ro7rs.~—l".vc1'y row contains one point which is distinguished from all otln-rs, viz., the. point at infinity. In two projccti'c i'ows, to the point I at inlinit__' in one corresponds a point 1’ in the other, and to the point J’ at infinity in the second corresponds a point J in the lir.~_.t. The points I’ and J are in general ﬁnite. If new A and It are any two points in the one, A’, 13’ the corresponding points in the other row, then Fig. 12. (ABJI) = (A’B’J’l’). AJ _ Al _ A’J’ _ ,i'i' Or. Jii ' in ’ 3'13" in" but, by § 17, ii _ A’J’ _ _] - Ill _ il"l3’ — ’ therefore the last equation changes into AJ A’l’ _ 1 J13 ' I'll’ or into AJ. A'I’ = DJ. ll'I', that is to say—— Theorem. —'l‘lic product of the distances of any two corresponding points in two projccti'c rows from the points which coi-rc.<pond to the points at infinity in the other is constant, viz., AJ. A’I' = 1.‘. Steiner has called this iiiinihcr k the Pozccr Q/' the corrr.-pomlcncr. § 39. ;S'im2'lar 1{uws.—Il' the points at infinity in two projective rows correspond so that I’ and J are at infinity, this result loses its nicaiiiiig. lut if A, B, C be any three points in one A’, B’, C’ the corresponding ones on the other row, we li-ave (ABCI) = (.’l}'C'I'), which reduces to AQ _ A'(' _ Y _t(!__ _ne cl; “ ’ ° -i'c' “ B’U” that is, corresponding segments are proportional. Conversely, if corresponding segments are proportional, then to the point at inﬁnity in one corresponds the point at inlinity in the other. ll‘ we call such rows sz'mz'Im-, we may state the result tliiis— Thcorem.—Two projective rows are similar if to the point at intinity in one corresponds the point at infinity in the otliH', and conversely, if two rows are similar then they are projective, and the points at infinity are corresponding points. From this the wcll-known propositions follow :— Two lines are cut proportionally (in siinilai- rows) by a Series of parallels. The rows are perspective, with centre of projection at inﬁnity. If two similar rows are placed parallel, then the lines joining homologous points pass through a common point. § 40. T/zeorem.——If two flat pencils he projective, then there exists in either mic single pair of lines at right angles to one another, such that the corresponding lines in the other pencil are again at right angles. To prove this, we place the pencils in perspective position (Fig. 13) hy making one ray coincident with its corresponding ray. (‘or- responding rays meet then on a line )7. And now we draw the circle which has its centre on p, and which passes through the