Xso that AX2≈ CD. BX. Then AX CD (AB-AX)≈ CD. AB - CD. AX≈ EF2 - CD. AX, EF being a mean proportional between AB and CD. Hence AX2+CD · AX≈EF2 ; :. AX2 + CD · AX + (} CD)2≈ EF2 + (} CD)2 = M2, say; hence AX+ { CD = M; ... AX = M − CD. Geometrically interpreted this means, find EF, a mean proportional between AB and CD; then find M, the hypotenuse of a rt. ▲ having EF and CD as arms; then AX = M - } CD. REMARK.-Note the close correspondence between this geometrical solution of a problem and the algebraic solution of a quadratic equation by completing the square. This method may be applied with facility to the solution of a great variety of problems concerning the sections of a line, a selection of which is appended below for exercise. This method has moreover the advantage of presenting simultaneously the solutions for both the internal and external section. Thus the problem just solved is one case of the more general problem: Divide a given line in internal or external section so that the square of one segment; etc. We found above that (AX+CD)2≈ M2, M being a line of determined length, but of direction either positive or negative; that is, it may be drawn in either of two opposite directions from a given point A. Hence we may take AX+ CD = M, whence AX= M— } CD, AX' = — M— } CD. If then AX, laid off to the right from A, gives the internal point of section, AX', laid off in the opposite direction, gives the external point of section, and AX22 ≈ CD.BX'. In the following extra examples, a to g, it is to be noted that divide is to be taken in the general sense of divide internally or externally. a. Divide a given line into segments such that the square of one segment shall be equivalent to the rectangle of the given line and the other segment. This important problem, the medial section, already solved synthetically in Art. 308, is given here as an example of the application of this method, which, it will be perceived, includes both the analysis and the synthesis of the problem. Let AX be one segment of the given line AB; then AB - AX is the other. By the given conditions, AX2≈ AB(AB—AX)≈ AB2 — AB · AX; .. AX2+ AB · AX≈ AB2; AX2+ AB • AX+(} AB)2≈§ AB2; · · AX+} AB = ± √31 B ; .. AX=(√5—1) 4B, AX'——(√5+1) AB, 2 2 The construction given in Art. 308, and that found in Euclid (II. 11) are merely different ways of interpreting the result AX=(√5−1)4B, To these we add a third construction, based upon Art. 363, and giving both AX and AX'. Bisect the given line AB in C, and produce CA to D, so that AD = AC. At D draw DE Land: = to DC or AB. From E as center, with radius equal to DB, describe an arc cutting AB in X and AD produced in X'. b. Divide a given line into segments such that the sum of their squares shall be equivalent to the rectangle of the given line and one of the segments. c. Divide a given line into segments such that the sum of their squares shall be equivalent to a given square. N.B. In this and similar examples rectilinear figure may be substituted for square, since a square can be found equivalent to any given rectilinear figure (360). d. Divide a given line into segments such that the difference of their squares shall be equivalent to the rectangle of the given line and a segment. N.B. The result obtained, AX=} AB, shows that there is but one, the internal, section. e. Divide a given line into segments such that the difference of their squares shall be equivalent to a given square. N.B. If AX be a segment of the given line, and CD the given square, we obtain the equality 2 AB⚫ AX≈ AB2 + CD2 ≈ M2, and AX is a third proportional to 2 AB and M. f. Divide a given line into segments such that the ratio of their squares is as m to n. g. Divide a given line into segments such that the rectangle of the segments shall be equivalent to a given square. N.B. On completing the square we obtain (AX — § A B)2 ≈ } AB2 — CD2≈ M2, say; whence AXAB± M, in which result M represents a possible result only when CD is not > AB; i.e., the given square must not be greater than the square of one half the given line. 434. Let P be the point through which passes the given chord AB. Draw the radius OPD, at P make CPD = Z BPD, and complete the chord CPE. Then CP: PE = PB: PA (Ex. 198, Ex. 197). 435. Let PXY be the required secant drawn from the given point P and meeting the circumference of O in X and Y, so that PX: XY= m: n. Draw a tangent PT. Since XY = "PX, and PX. PY≈ PT2 ̧m m (303), PX(PX+PX) + " + " PX2 ~ PT2 ; · · m or PX: PT = √m: √m+n, and PX can be found by Art. 363. Y X A B 436. Let P be the point in the arc whose chord is AB, and PXY the required chord bisected by AB in X. Join OX, OP; OX is to PY (177), and OXP is a right angle; .. a semicircumference described on OP as diameter will pass through X. Hence join OP, on OP as diameter describe a semicircumference; etc. 437. Let OA, OB be the radii perpendicular to each other. Join AB, and produce each way to C and D, so that BC, AD are each equal to AB. Draw OC, OD, cutting the circumference in X and Y respectively, and join XY. Since 20AB LOBA (68), ≤OAD = ZOBC; also OA, AD=0B, BC respectively; ... ▲ OAD=▲ OBC; ... ODOC. But OY OX; ... XY is parallel to CD (276); ... XZ = ZV = VY (282), since CB = BA = AD (Const.). Y C X = 438. Let XY be the required chord | to AB in 0. Draw OC I to XY, meeting AB, in D; draw XE 1 to DB. DE: DB CX: DB = 2 = produced if necessary, Since DE = CX (136), CX: 2 DB = XY: AB. Hence divide DB in E so that DE: DB is the given ratio. Draw EX to DB; etc. 439. Let XY be the secant divided at P in the given ratio. Join 00'; draw OA, to A and divide 00' in C in the given ratio; join CP, and through P draw XPY to CP; etc. 440. Let DG be the required square. Draw the altitude AH to BC, and AK parallel to BC. Join CD and produce to meet AK in K; draw KL perpendicular to CB produced. By similar triangles, AK: Α K D B EH G DF = KC: DC, and KL: DE = KC: DC; :. AK: KL = DF: DE. Hence draw AH perpendicular to BC, and AK parallel to BC and equal to AH; draw KC cutting AB in D; DE is a side of the required square. E Y C FB 441. Let XF be the required square in the segment whose chord is AB. Bisect AB in C, and at A draw AD perpendicular to AB; join CX and produce to meet AD in D. By similar triangles, DA: AC=XY: CY. But XY= YF=2 CY; .. DA=2 AC=AB. Hence at A draw AD perpendicular to AB and equal to AB; bisect AB in C, and draw CD cutting the circumference in X; draw XY perpendicular to AB; XY is a side of the required square. 442. Let XF be the required rectangle inscribed in the semicircle whose diameter is AB. At A draw AD 1 to AB, bisect AB in C, join CX, and produce to meet AD in D. By similar triangles, AD: AC XY: YC; .. AD: 2 AC or AB XY: 2 YC or YF. Hence at A draw a perpendicular to AB, and lay off AD so that AD: AB is the ratio of the sides of the given rectangle. Bisect AB in C, and join CD, cutting the circumference in X; draw XY1 to AB, and XE || to AB; etc. = D AY E FB 443. Let ABC be the given triangle and DEF the given circle. At any point D of the circumference draw a tangent GDH, at D make GDF = LB, and ZHDE = ¿C. Join EF; then as E = ZGDF=LB, and / F=ZHDE=LC, A DEF is similar to AABC. 444. Let ABC be the given triangle, and DEF the given circle. Draw radii OD, OE, making an DOE equal to the supplement of A, and draw a third radius OF making FOD equal to the supplement of ZB. Through D, E, F, draw tangents intersecting in H, K, L. The & DOE, DOF being respectively the supplements of H and K, as well as of A and B, we have 2 H= ZA, 2 K = ZB; etc. K F 445. Let DEFG be the required par'm inscribed in ▲ ABC. From A draw AH || to GD, and AK || to BC; draw CG, and produce to meet AK in K. By similar A, AK: GF = KC: GC, and KL: GD = KC: GC; AK: : KL =GF: GD (249). Hence draw AH making an angle with BC equal to an angle of the given par'm, draw AK || to BC so that AK: KL is the ratio of two adjacent sides of the par'm; draw CK meeting AB in G, draw GF || to BC, and GD, FE, || to AH; etc. 446. Let ABC be the required triangle, having its base BC, and AD the bisector of its vertical angle, given, as also its vertical angle. Describe B A D E Ᏼ Ꭰ Ꮋ E C about ABC a circumference ABED, produce AD to E, and join EB. Since ZBAE=/CAE, E is the mid point of arc BEC; and since Z BAE CAE = LDBE (266), and ▲ BED is common, ▲ ABE is similar to ABDE; ... AE: EB = EB: DE; ... AE. DE~ EB2; DE(DE + AD)≈ EB2; i.e., DE2 + AD · DE ≈ EB2; ... DE2 + AD · DE + (≥ AD)2 ≈ EB2 + (} AD)2 ≈ M2, say; :. DE + AD = M; ... DE + AD = M + 1⁄2 AD. Hence upon BC as base, describe a segment BAC containing an angle equal to the given angle (311), and complete the circumference by arc BEC. Bisect BEC in E, and join EB. Find M, the hypotenuse of a right triangle, whose arms are equal to AD and EB respectively; then with E as center and radius = MAD, describe an arc cutting the circumference in A; join AB, AC, AD; etc. REMARK. Two exercises connected with this one, and leading up to it, were accidentally omitted from among the theorems. They are: (a) The bisectors of all angles inscribed in a given segment meet in a point. This point, it will be seen, is the mid point of the intercepted arc. (b) If from the mid point of a given arc any chord be drawn cutting the chord of that arc, the rectangle of the chord thus drawn and its segment intercepted between the arc and its chord, is constant. Thus we proved above that AE DE EB2, a constant. LOCI, p. 164. 447. The perpendicular at the mid point of the line joining the given points being the locus of all points equidistant from those points (213), |