Propositions Selected from a Paper on the Division of Right Lines, Surfaces, and Solids. By James Glenie, A. M. of the University of Edinburgh. Communicated by the Astronomer Royal

III. Propositions selected from a Paper on the Division of Right Lines, Surfaces, and Solids. By James Glenie, A. M. of the University of Edinburgh. Communicated by the Astronomer Royal. R. June, 1775. PROPOSITION I. THEOREM. If from the angles at the base of any right-lined triangle, right lines be drawn to the alternate angles of rhombi, described upon the opposite sides, and applied reciprocally to the sides produced; and from the vertex, through the intersection of these lines, a right line be drawn to meet the base: the segments of the base, made thereby, will have to each other the duplicate proportion of the sides. Let ACB be any right lined triangle. Let AFEC, CDGB be rhombi, on any two sides AC, CB of this triangle, applied respectively to CB, AC, produced: from the alternate angles EFA, DGB, of which let FA, GA, be right lines drawn to the angles at the base, or third side, AB. Then, if through the intersection O of these these lines, a right line \( \text{col} \) be drawn from the vertex \( c \) to meet the base \( ab \); the segments \( al, lb \), of the base made thereby, will have to each other the duplicate proportion of the sides \( ac, cb \). For through the vertex \( c \) let there be a right line drawn parallel to \( ab \), to meet \( bf, ag \) produced, if necessary. Then, since the triangles \( cqh, cpi \), are respectively equiangular to the triangles \( aqb, apb \) (15. and 29. E. i.); the proportions of \( ch \) to \( ab \) and of \( ab \) to \( ic \) are respectively equal to the proportions of \( cq \) to \( qb \) and of \( ap \) to \( pc \) (4. E. 6.). But the proportion of \( ch \) to \( ic \) is compounded of the proportion of \( ch \) to \( ab \), and of \( ab \) to \( ic \); and consequently is equal to the proportion compounded of the proportions of \( cq \) to \( qb \), and of \( ap \) to \( pc \). And, since the triangles \( acq, apf \), are respectively equiangular to the triangles \( bqg, bpc \) (15. and 29. E. i.); the proportions of \( cq \) to \( qb \) and of \( ap \) to \( pc \) are each equal to the proportion of \( ac \) to \( cb \) (4. E. 6.); and when compounded are equal to the duplicate proportion of \( ac \) to \( cb \). Wherefore the proportion of \( ch \) to \( ic \), which hath been shewn to be equal to the proportion compounded of the proportions of \( cq \) to \( qb \) and of \( ap \) to \( pc \), is also equal to the duplicate propor- tion of AC to CB (II. E. 5.). But, since the triangles COH, COI, are respectively equiangular to the triangles AOL, LOB, the proportion of CH to IC is equal to the proportion of AL to LB (4. E. 6. and 16. E. 5.). Therefore the proportion of AL to LB is equal to the duplicate proportion of AC to CB (II. E. 5.). Q.E.D. COR. I. If the triangle be isosceles, the right line drawn from the vertex to the base is perpendicular thereto, and the segments of the base are equal to each other. COR. II. When the triangle is right-angled, the line drawn from the vertex to the base is always perpendicular to it (as appears from 8. E. 6. and its cor.); and the rhombi become squares on the sides comprehending the right angle. COR. III. The segments of the sides adjacent to the base, are respectively third proportionals to the sum of the sides, and the sides themselves. COR. IV. The segments of the sides adjacent to the vertex are equal to each other, and each of them is a fourth proportional to the sum of the sides, and the sides themselves (a). COR. (a) And it may be added, a mean in proportion between the two segments adjacent to the base. For if a right line AB be any how divided in C, and from A D C E B the two segments CA, CB, third proportionals to the whole line and each segment respectively, CD, CE, be taken away, the remainders AD, EB, are equal, and each is a mean in proportion between the two CD, CE. For because AB : AC = AC : CD; therefore, COR. V. The segments of the base are proportional to the segments of the sides, which are adjacent to them. PROPOSITION II. Let there be any two right lines given. There is an angle which may be made by these lines; such that if, from their extremities which do not meet, right lines be drawn to the alternate angles of rhombi described on them, and reciprocally applied to them when produced; and from the said angle through the intersection of these lines, a right line be drawn to meet the right line joining the said extremities; the segments of this line made thereby, shall be respectively equal to the adjacent segments of the given lines. Therefore, by conversion, \( AB : BC = CA : AD \). Again, because \( AB : BC = BC : CE \), by conversion \( AB : AC = BC : BE \); and by permutation \( AB : BC = AC : BE \). Therefore \( AC : BE = AC : AD \). Therefore \( AD \) and \( BE \) are equal. I say, moreover, that each of the two equal lines \( AD, BE \), is a mean in proportion between the two \( CD, CE \). For because \( BA : AC = AC : CD \), by division \( BC : CA = AD : DC \). Again, because \( BA : BC = BC : CE \), converting and dividing \( BC : CA = CE : EB \). Therefore, \( CE : EB = AD : DC \). Q.E.D. S. HORSLEY. Let AC, CB, be any two given right lines, and let CD in AC produced be equal to CB. On AD describe a semi-circle; draw CN at right angles to AD, and equal to CD (II. E. 1.); join A, N, and apply a right line AM in the semi-circle equal to AN (I. E. 4.). From the point M draw the right line MS at right angles to AD (I. E. 4.) Make a triangle ACB, having its sides equal to AC, AS, and CB (by 22. E. 1.); and ACB is the angle required to be found; and the segments AL, LB, of the right line AB joining the extremities A and B of the given lines are respectively equal to the segments AP, BQ, of the given lines, which are adjacent to them. For the square on BC hath to the square on AC the duplicate proportion of BC to AC (cor. I. to 20. E. 6.); that is, the proportion of BL to LA (prop. I.). Wherefore the squares on AC, CB; that is, the square on AN or AM (47. E. 1.) hath to the square on AC the proportion of AB to AL (I8. E. 5.). But the square on AN or AM is equal to the rectangle contained by AD, AS (8. and 17. E. 6.). Wherefore the rectangle contained by AD, AS, hath to the square on AC, the proportion of AB to AL; that is, the proportion of the rectangle contained by AB, AS to the rectangle contained by AL, AS (I. E. 6.). Consequently, the proportion of the rectangle AD, AS, to the rectangle rectangle $AB$, $AS$; that is, the proportion of $AD$ to $AB$ (I. E. 6.) is equal to the proportion of the square on $AC$ to the rectangle $AL$, $AS$ (16. E. 5.) But $AD$ hath to $AC$ the proportion of $AC$ to $AP$ (cor. 3. to prop. I.). Therefore, the rectangle $AD$, $AC$, hath to the square on $AC$ the proportion of the rectangle $AC$, $AS$, to the rectangle $AP$, $AS$ (I. E. 6.). Therefore $AD$ hath to $AS$ or $AB$ the proportion of the square on $AC$ to the rectangle $AP$, $AS$ (16. E. 5. and I. E. 6.). Hence the rectangle $AL$, $AS$, is equal to the rectangle $AP$, $AS$ (II. and 16. E. 5.), and $AL$, $AP$, are consequently equal. But as $AL$ to $AP$, so is $BL$ to $BQ$ (by cor. 5. prop. I.). Therefore $BL$, $BQ$, are likewise equal. Q.E.D. PROPOSITION III. PROBLEM I. To multiply the square of a given finite right line by any number. UPON an indefinite right line AP set off the given right line AB; draw BC at right angles to AP and equal to AB; and from A through C draw an indefinite right line AQ. Take AD equal to AC, and draw DE parallel to BC; AF equal to AE; and draw FG parallel to BC, and so on. Then it appears (from 47. E. i.), that the square of AC is is equal to the square of \( AB \) multiplied by 2; the square of \( AE \) equal to the square of \( AD \) or \( AC \) multiplied by 2; that is, equal to the square of \( AB \) multiplied by 4, and so on. Thus the squares of \( AC, AE, AG, AI, AL, \&c. \) are respectively equal to the square of \( AB \) multiplied by the terms of the following series 2, 4, 8, 16, 32, 64, &c. where the sixty-third term gives the square of \( AB \) multiplied by the last term of Sessa's Series for the Chessboard. If \( CX \) be drawn parallel to \( AP \), the squares of \( AA, AB, AC, AD, \&c. \) will be respectively equal to the square of \( AB \) multiplied by 3, 5, 9, 17, 33, 65, 129, &c. Also if \( AG \) be taken equal to \( AA \), and \( GE \) be drawn parallel to \( BC \), and this be repeated, the squares of \( AE, \&c. \) will be equal respectively to the square of \( AB \) multiplied by 6, 12, 24, 48, &c. And the squares on \( AO, \&c. \) will be equal to square on \( AB \) multiplied by 4, 7, 13, 25, 49, &c. In like manner, if \( AM \) be taken equal to \( AB \), and \( MN \) be drawn parallel to \( BC \), the squares on \( AN, \&c. \) will be equal respectively to the square on \( AB \) multiplied by 10, 20, 40, 80, 160, &c. And the squares on \( AS, \&c. \) will be equal respectively to the square on \( AB \) multiplied by the terms of the following series: 6, 11, 21, 41, 81, 161, &c. In the same way, if right lines be drawn from \( E, E, G, N, L, \&c. \) there will arise numberless other series. And if \( BC \) be taken equal to \( AB \) multiplied by any number, surd, fractional, or mixed, there will be obtained a great variety of series, consisting respectively of terms, which are are surd, fractional, or mixed. And by dividing BC, DE, GE, FG, MN, HI, KL, in different ways, according to pleasure, we may apply the same method to fractional numbers, without altering the magnitude of BC. Thus, if BC be bisected, and a right line be drawn through the point of bisection parallel to AP, there will be found right lines, the squares on which are respectively equal to the square on AB multiplied by a great number of fractions, having four for their common denominator, and so on. PROPOSITION IV. PROBLEM II. To find a right line, the square on which shall be equal to the square on a given right line, divided by any number. IF, using the figure of the immediately preceding problem, we suppose the given right line to be denoted by AL, the squares on AK, AH, AF, AD, AB, &c. will respectively be equal to the square on AL multiplied by $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{16}$, $\frac{1}{32}$, $\frac{1}{64}$, $\frac{1}{128}$, $\frac{1}{256}$, $\frac{1}{512}$, $\frac{1}{1024}$, &c. or divided by 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, &c.; and so on for other numbers, whole, surd, fractional, or mixed. PROPOSITION V. PROBLEM III. To cut off from a given right line a part expressed by any odd number. LET \( AB \) be the given right line. At right angles to it, at one of its extremities \( B \), draw an indefinite right line \( BE \). Let \( n \) be the given odd number, which expresses the part of \( AB \) which is to be cut off. Take \( BC \) such a right line (prob. 8.), that the square on it shall be equal to the square on \( AB \), multiplied by the number \( \frac{n-1}{2} \). Draw \( CL \) as in the first theorem, and take \( LS \) equal to \( LB \). I say, as is that part of \( AB \), which is expressed by the odd number \( n \). For the square on \( AC \) being equal to the squares on \( AB \), \( BC \), is equal to the square on \( AB \) multiplied by the number \( \frac{n-1}{2} + 1 \), or \( \frac{n+1}{2} \). Therefore it appears (from prop. 1. and cor. 1. to 20. E. 6.), that \( AL \) is to \( LB \) as \( \frac{n-1}{2} + 1 \) to \( \frac{n-1}{2} \). Consequently, as is equal to the part required. Q.E.F. Thus, Division of Right Lines, Surfaces, and Solids. Thus, if the square on BC be supposed successively equal to the square on AB multiplied by the terms of the series 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, &c. the numbers of the several parts denoted by AS, will be 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, &c. which series comprehends all odd numbers after 9, and might have begun from 3 had the other series begun from 1. PROPOSITION VI. PROBLEM IV. To cut off from a given right line a part expressed by any even number. Let $m$ denote any even number in general. Draw any indefinite right line $BH$, and at right angles to it another $BE$. On $BE$ set off the given right line $BA$, and from $A$, with the distance equal to a right line, the square on which is equal to the square on $AB$ multiplied by the number number \( m - 1 \), intersect \( BH \) in some point \( C \). From the vertex \( A \) of the triangle \( BAC \) draw \( AL \) as was directed in prop. 1. and draw \( LS \) parallel to \( CA \). I say \( BL \) is such a part of \( BC \) as is expressed by the number \( m \); and that \( BS \) is the same part of \( AB \). For it appears (from prop. 1.) that \( BL \) is to \( CL \) as 1 to \( m - 1 \). Wherefore \( BL \) is the \( m^{th} \) part of \( BC \). And \( BS \) is the same part of \( AB \) that \( BL \) is of \( BC \) (4. E. 6.) Thus if the square on \( AC \) be successively denoted by the square on \( AB \) multiplied by 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, &c. \( BS \) will be successively such a part of \( AB \) as is expressed by 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, &c. **Proposition VII. Theorem II.** If from the angles at the base of any right lined triangle, right lines be drawn to the alternate angles of rhombi described on the other two sides, and reciprocally applied to them produced, and through the intersection of these lines, a right line be drawn from the vertex to the base; the rectangle contained by the sines of the angles at the extremities of one of the sides, will be equal to the rectangle contained by the sines of the angles at the extremities of the other; and the parallelepiped contained by the sines of the \((b)\) The author means, that \( \sin. ACL \times \sin. CAL = \sin. BCL \times \sin. CBL \) (see fig. prop. 1.). For \( \sin. ACL : \sin. L = AL : AC \), and \( \sin. L : \sin. BCL = BC : BL \). Take \( N \), a third in proportion to \( AL, AC \). Then, because \( AC^2 : BC^2 = AL : LB \), \( N \) will likewise the angles of one of those triangles, into which the original one is divided by the said line drawn from the vertex, will be equal to the parallelepiped contained by the sides of the angles of the other. COR. The two triangles, adjacent to the segments of the base, have to each other the proportion of the two adjacent to the sides containing the vertical angle, or the proportion of the two into which the original triangle is divided; and any one of these pairs of triangles are as similar figures described on the sides, being as the segments of the base, which have to each other the duplicate proportion of the sides. PROPOSITION VIII. THEOREM III. If from the angles at the hypotenuse of any right angled right lined triangle, right lines be drawn to the alternate angles of squares described on the sides containing the right angle, and from the point where the right line drawn from the right angle, through their intersection, meets the hypotenuse, right lines be drawn to the points, where these lines meet the sides; the lines so drawn will likewise be a third in proportion to \( l_b, b_c \). Hence, \( \sin. acl : \sin. l = ac : n \), and \( \sin. l : \sin. bcl = n : bc \). Ex aequo perturbate \( \sin. acl : \sin. bcl = ac : bc \). But \( \sin. cbl : \sin. cal = ac : bc \). Therefore, \( \sin. acl : \sin. bcl = cbl : \sin. cal \), and \( \sin. acl \times \sin. cal = \sin. bcl \times \sin. cbl \). Q.E.D. And hence, \( \sin. l \times \sin. acl \times \sin. cal = \sin. l \times \sin. bcl \times \sin. cbl \), which is the second branch of the proposition. S. HORSLEY. make equal angles with the hypotenuse, and the right line drawn from the right angle to meet it; and will likewise have to each other the proportion of the sides containing the right angle. COR. (c) For in the triangle ACB, having the angle at C right, construct the figure as in the first proposition, and join LP, LQ. The author affirms, that the angles PLA, QLB, and also the angles CLP, QLC, are equal, and that PL : LQ = CA : CB. Produce LP, LQ, till they meet the right line IH in M, N. Now, because MN, AB, are parallel, therefore AL : MC = AP : PC, and CN : LB = CQ (or PC) : QB. Therefore, AL × CN : MC × LB = AP : QB. But AP : QB = AL : LB (by theor. I. cor. 5.). Therefore, AL × CN : MC × LB = AL : LB. Therefore, CN and CM are equal. But because the angle ACB is right, CL is perpendicular to AB (by theor. I. cor. 2.), and consequently to MN. Therefore the angles MCL, NCL, are right, and MC being equal to CN, and CL common to the two triangles MCL, NCL, ML will be equal to LN, and the angles MLC, CML, equal to NLC, CNL, respectively. Hence it is evident, that the angles CLP, CLQ, are equal, and each is half a right angle; and likewise PLA, QLB, are equal, and each half a right angle. Further, COR. I. The alternate triangles of those four, which have their vertices in the point, where the right line drawn from the right angle meets the hypotenuse, are similar, and have to each other the proportion of the segments of the hypotenuse, or the duplicate proportion of the sides containing the right angle. COR. II. Either pair of the adjacent triangles lying on different sides of the right line drawn from the right angle, and having their vertices in the intersection of the right lines drawn from the angles at the hypotenuse, have to each other the proportion of the alternate triangles, having their vertices in the intersection of the first-mentioned line and the hypotenuse. COR. III. The trapezium or quadrilateral figure formed by the segments of the sides adjacent to the right angle, and the right lines joining their extremities with the intersection of the hypotenuse, and the right line drawn from the right angle to meet it, is capable of being inscribed in a circle; and is divided at the intersection of right lines drawn from the angles at the hypotenuse to the alternate angles of squares, described on the sides containing the right angle, into triangles which are proportional to one another, and when taken two by two, as Further, the angles, \( \text{PLA}, \text{CQL} \), are equal, each being half a right angle. And the angles \( \text{PAL}, \text{LCQ} \), are equal (by 8. Elem.6.). Therefore the angles \( \text{APL}, \text{CQL} \), will be equal, and the triangles \( \text{APL}, \text{CQL} \), similar, and the sides subtending the equal angles proportional. Therefore, \( \text{PL : LA = QL : LC} \). By permutation, \( \text{LP : LQ = LA : LC} \). But \( \text{LA : LC = AC : CB} \) (by 8. Elem. 6.). Therefore, \( \text{LP : LQ = AC : CB} \). Q.E.D. S. HORSLEY. they lie adjacent on different sides of the diagonal, are proportional to the unequal sides of the trapezium, and to the two triangles into which the diagonal divides it. For \( \text{OPL} : \text{OPC} = \text{LO} : \text{OC} = \text{OQL} : \text{OQC} \). Therefore, \( \text{OPL} : \text{OQL} = \text{OPC} : \text{OQC} = \text{LPC} : \text{LQC} = \text{LP} : \text{LQ} \). **PROPOSITION IX. THEOREM IV.** *If from the angles at the base of any right lined triangle, right lines be drawn to the alternate angles of rhomboids described on the other two sides, and reciprocally applied to them produced, a right line drawn from the vertex through the intersection of these lines will cut the base into two parts, having to each other the proportion compounded of the proportion of the sides, and of the proportion of the other two lines comprehending the rhomboids.* I SHALL subjoin a demonstration of this theorem, since the first proposition in these papers is only a particular case of it. Let the triangle be \( \triangle ACB \), the base \( AB \), the rhomboids \( ACEF, CC'DG \); and let the right lines \( BF, AG \), be drawn. Then, if from the vertex \( C \) through their intersection \( O \), a right line \( CL \) be drawn to meet the base, the segments \( AL, LB \), thereof will have to each other the proportion compounded of the proportions of \( AC \) to \( CB \), and of \( CE \) to \( CD \). For through the vertex \( C \) let a right line \( ICH \) be drawn parallel to \( AB \), to meet \( BF, AG \), produced, if necessary. Then, since the triangles \( CQH, CPI \), are respectively equiangular to the triangles \( AQB, APB \) (15. and 29. E. i.), the proportions of \( CH \) to \( AB \), and of \( AB \) to \( IC \) are respectively equal to the proportions of \( CQ \) to \( QB \), and of \( AP \) to \( PC \) (4. E. 6.). But the proportion of \( CH \) to \( IC \) is compounded of the proportions of \( CH \) to \( AB \) and of \( AB \) to \( IC \); and consequently is equal to a proportion compounded of the proportions of \( CQ \) to \( QB \) and of \( AP \) to \( PC \). And since the triangles \( ACQ, APF \), are respectively equiangular to the triangles BQG, BPC (15. and 29. E. 1.); the proportions of CQ to QB, and of AP to PC, are respectively equal to the proportions of AC to BG or CD, and of AE or CE to CB. Wherefore the proportion of CH to IC is equal to the proportion compounded of the proportions of AC to CD and of CE to CB, or of AC to CB and of CE to CD. But since the triangles COH, COI, are respectively equiangular to the triangles AOL, LOB, the proportion of CH to IC is equal to the proportion of AL to LB (4. E. 6.). Therefore the proportion of AL to LB is equal to the proportion compounded of the proportions of AC to CB and of CE to CD. Q. E. D. SCHOLIUM. If CE, CD, be equal to each other, AL hath to LB the proportion of AC to CB, and CL bisects the angle ACB; if CE have to CD the inverse proportion of AC to CB, AL is equal to LB; if CE have to CD the proportion of AC to CB, AL hath to LB the duplicate proportion of AC to CB; and universally, if CE have to CD any multiplicate proportion, n, of AC to CB, AL hath to LB such a multiplicate proportion of AC to CB as is expressed by the number n+1. And if CE have to CD any multiplicate proportion m of CB to AC, AL will have to LB such a multiplicate proportion of CB to AC, as is expressed by the number m−1.