940. COR.-Conversely, If three points be taken in the sides of a triangle (as a, b, c) such that the product of three non-adjacent segments equals the product of the other three, the points are in the same straight line. For, passing a line through a and b, let it cut the third side in c'. Then, by the proposition, aA × b°C × c'B = aC × bB × c'A. But, by hypothesis aA × bC x cB = aC × bB × cA. Whence = and c and c' must coincide. св CA c'B C'A' SCH.-This theorem is known among mathematicians as The Ptolemaic Theorem, and is usually attributed to Claudius Ptolemy, an Egyptian mathematician and philosopher who flourished in Alexandria during the first half of the second century. But it is thought to be more properly due to Menelaus, who lived a century before Ptolemy. 941. Theo-The three angle-transversals* of a triangle, passing through a common point, divide the sides into segments such that the product of three non-adjacent segments equals the product of the other three. DEM.-From the triangle ACe cut by the transversal aB, we have aA × CO × cB = AB × aC × Oc; and from CBC cut by bА, Oc × ¿C × AB = CO × bB × cA. Multiplying, we obtain aA x bC xc BaC × bB × cA. 942. COR. 1.-Conversely, If the three angletransversals of a triangle divide the sides into segments such that the product of three non-adjacent segments equals the product of the other three, the transversals pass through a common point. O B FIG. 484. B For, the sides being divided at a, b, and c, so that aА × b°C × cB = aC × bB × CA, draw Ce, and Ab, and let O be their intersection. Now, let a' be the point in which BO cuts AC. Then, by the proposition, a'A × bC × cB = a'C × bB × CA. aA ac Whence = a'A a'C' and a and a' coincide. 943. COR. 2.—If any one of the sides is bisected, the line joining the other points of division is parallel to this side. For, let bCbB. Then a A × b°C × cB = aC × bB × cA, becomes αA × cB = aC × cA; or Aa : aC :: Ac: cB. QUERY.-How does this apply to the second figure? 944. COR. 3.—If the line joining two points of division is parallel to the third side, the latter side is bisected. * The transversals passing through the angles. For, if ab is parallel to AB, aC : bC :: aA :b3, whence aC × bB = bC × ¤Â ̧ And, since a▲ × bC × cB = aC xbB xcA, CA: = CB. 945. We will now give a few problems to illustrate the use of the theory of transversals. 946. Prob.-To show that the medial lines of a triangle pass through a common point. с FIG. 485. B SOLUTION. Since aA = : aC, bC = bB, and cB = cA, by multiplying, we have aА × b°C × cB = aC × bB x cA; whence by the last corollary these transversals pass through a common point. 947. Prob.—To show that the bisectors of the angles of a triangle pass through a common point. SOLUTION. In the last figure let aB, bA, cC be the bisectors. «А AB ¿C AC cB CB Then ас = = = aA × bCx cB = 1, or CB' B AB' CA AC; multiplying, aC x bB x CA aA × bC × cBaC x bB × CA. Therefore these transversals pass through a common point. 948. Prob.-To show that the altitudes of a triangle pass through a common point. SUG's. In the last figure, if aB, bA, cC, were the perpendiculars, there would a A АО С Со св OB be three pairs of similar triangles giving BO' CA AO' aC whence, as in the last. = ᏓᏴ = = CO 949. Prob.-To show that the angle-transversals terminating in the points of tangency of the sides of the triangle with its inscribed circle, pass through a common point. SUG's. In the last figure, if a, b, c were the points of tangency we should have aA CA, ¿C = aC, cB =bB; whence aA × b°C × cB = aC × bB × cA. Which shows that the transversals pass through a common point. = 950. Theo-If two sides of a triangle are divided proportion D B ally, starting from the vertex, the angle-transversals from the extremities of the other side to the corresponding points of division, intersect in the medial line to this third side. DEM.-Since AC and CB are divided proportionally at a and a', aА × a'C = aC × a'B; and as DB = DA, αA x a'C × DB = aC x a'B x DA, the angle-transversals Aa', Ba intersect in CD. The same may be shown of any other angle-transversals from A and B, dividing CB and CA proportionally. FIG. 486. 951. COR.-In any trapezoid the transversal passing through the intersection of the diagonals, and the intersection of the nonparallel sides, bisects the parallel sides. SUG.-Joining aa' in the last figure, CD is such a transversal. The student will readily see the connection with the proposition. 952. Prob.-Through a given point to draw a line which shall meet two given lines at their intersection in an invisible, inaccessible point. SOLUTION.-Let Mm, Nn be the two given lines which meet in the invisible, inaccessible point S, and P the given point through which a line is to be located which will meet Mm, Nn in S. Through P draw any convenient transversal, as BF, and any other meeting this, as AF. Now, considering MS as a transversal of the triangle CDF, we have M AF x BC x SD = AD x BF x SC; whence SD AD × BF = SC AF x BC But, HD being drawn HD SD AD x BF parallel to BF, we have = = B P H FIG. 487. N whence HD is known, as AD, BF, PC, AF, BC can be measured. The points P and H determine the required line. 953. DEF.-The Complete Quadrilateral is the figure formed by four lines meeting in six points. The complete quadrilateral has three diagonals. ILL-ABCDEF is a complete quadrilateral, and its diagonals are CF, BD, and AE, the latter being spoken of as the third or exterior diagonal. A FIG. 488. E 954. Theo.-The middle points of the three diagonals of a complete quadrilateral are in the same straight line. DEM.-m, n, o being the centres of the diagonals of the complete quadrilateral, in the preceding figure, are in the same straight line. Bisect the sides of the triangle FDE, as at I, N, L, and draw IN, IL, LN. Since IN is parallel to BE, and bisects DF and DE, it also bisects DB (?) and hence passes through n. For like reasons I passes through m, and LN through o. Now, AC being a transversal of the triangle FDE gives CD × BE × AF = CE × BF × AD. Therefore, noticing that CD ml, = BE = nN, †AF = oL, ‡CE=mL, †BF = nl, and †AD = oN, we have ml × nNo × oL = mL × n¡ × oN. Hence these three points m, n, o lie in a transversal to the triangle ILN. SECTION V. HARMONIC PROPORTION AND HARMONIC PENCILS. 955. DEF.-Three quantities are in Harmonic Proportion when the difference between the first and second is to the difference between the second and third, as the first is to the third. 4:43::6:3. In gen ILL.-6, 4, 3 are in harmonic proportion, since 6 eral, a, b, c are in harmonic proportion, if a-b: b c::a: c. 956. Theo.—If a given line be divided internally and externally in the same geometric ratio, the distance between the points of division is a harmonic mean between the distances of the extremities of the given line from the point not included between them. A Let AB be the given line; and let O and O' be so taken that FIG. 490. B AO: BO :: AO': BO'; then is 00' a harmonic mean between AO' and BO'. For AO = AO' — 00′, and BO = 00′ — BO′; whence AO', OO', and BO' are such that AO - 00': 00' -BO' :: AO: BO', that is, they are in harmonic proportion. 957. COR. 1.—AO, AB, and AO' are in harmonic proportion, i.e., AB is a harmonic mean between AO and AO'. For AB AO (— BO): AO' — AB (= BO') : : AO: AO', 958. COR. 2.-When AO, OO', BO are in harmonic proportion, 959. COR. 3.-Conversely, When a line is divided into three parts such that the rectangle of the extreme parts equals the rectangle of the mean part into the whole line, the line is divided harmonically. = Thus, let AO' be the line, and AO' × BO' — BO × AO'; then AO: BO :: AO' : BO', whence, by the proposition, OO' is a harmonic mean between AO' and BOʻ. DEF. The points O and o' are called Harmonic Conjugates. 960. Theo.-If two lines be drawn, one bisecting the interior and the other the adjacent exterior angle of a triangle, and meeting the op * posite side, they divide this line harmonically. SUG-By means of (358, 359, PART II.) the student will be enabled to establish the relation AO: BO :: AO': BO', whence, O B FIG. 491. by the last proposition, AO', OO', BO' are in harmonic proportion. 961. Prob.-Given a right line to locate two harmonic conjugate points. Taking this by diviThe first three terms SOLUTION.-Let AB be the line. O may be taken at pleasure between A and B. We are then to find O', so that AO: BO :: AO': CO'. sion, we have AO – CO : BO :: AO' — BO' (= AB): BO'. being kuown, the other can be constructed. Or, we may first locate O' at pleasure, and then find O. 962. Theo.-If from the given point C in a line the distances O B CO, CB, CO' be taken in the same direction, so that CO × CO'=CB2; and if CA: = CB be taken in the opposite di rection, AO' will be divided harmonically at O and B. DEM. - From CO × CO' = CB2, we readily write CO : CB :: CB: CO'. CB + CO (= AO) : CB – CO (= BO) :: CO + CB (= AO) : CO – CB (= BO′). 963. COR. 1.-Conversely, If a line AO' be cut harmonically at O and B, and either of the harmonic means be bisected, as AB at C, the three segments CO, CB, CO' will be in geometric proportion. For, since AO': BO' :: AO: BO, AO' + BO' : AO' — BO' :: AO + BO : AO-BO, or 2CO': 2CB :: 2CB: 2CO, and CO': CB :: CB: CO. 964. COR. 2.-In a given line, as AB, as O approaches the centre C, O' recedes, and when O is at C, O' is at infinity, since CO' = CB CO 965. Theo.-The geometric mean between two lines is also the geometric mean between their arithmetic and harmonic means. DEM.-Let AO' and BO' be the two lines. On their difference, AB, draw a semicircle, draw the tangent O'T and let fall the perpendicular TO. Then O and O' are harmonic conjugates, since CO × CO' = CB2 (?), CO' is the A C O B FIG. 493. * The bisector of the exterior angle meets the side produced; but in higher geometry, as it is: always understood that lines are indefinite unless limited by hypothesis, such specifications aredeemed unnecessary. |