that perpendicular is the locus of the centers of all circumferences passing through those points. 448. The line that passes through the center of the given circle and through the given point, will pass through the center of every circle that is tangent to the given circle at that point (199); hence that line is the required locus. 449. The central distance of tangent circles equals the sum or difference of their radii; hence the locus is two circumferences concentric with the given circle, with radii equal to the sum and difference of the radius of the given circle and the given radius. 450. The perpendicular to the given line at the given point passes through the center of every circle that is tangent to the line at that point (192); hence that perpendicular is the required locus. 451. The bisectors of the vertical angles formed by intersecting lines constitute the locus of points equidistant from those lines (216); hence they constitute the required locus. 452. Let P be a point in the locus, from which tangents PA, PB, of a given length, are drawn to O 0. Join OA, OB, OP. Since OA, PA are constants, OP is also constant. Hence the circum ference concentric with the given circle and having a radius equal to the hypotenuse of the right triangle having OA, PA as arms, is the required locus. 453. Let AB be a chord passing through the given point P in O 0. Join OP, and draw OC 1 to AB. OCP is a right triangle, and C is a point in the locus. Hence the circumference described on OP as diameter is the required locus. 454. Let P be the given point without O 0. Join PO, and draw radius OA 1 to PO; join PA. Through B, the mid point of PO, draw BC to OA to meet PA in C. The circumference described from B as center with radius BC, is the required locus. Draw any secant PD; join OD, and through B draw BE || to OD. Then (147) ED=EP and BE=} OD=} OA= BC (148); .. E, the mid point of PD, is a point in the circumference of B. It may B easily be shown that that circumference bisects also the external segments of the secants. 455. On the given base describe a segment containing an angle equal to the given angle; the arc of this segment is the required locus. 456. Let P, Q be the given points, and M a point in the required N M S P R locus. Draw PMN, PQ, and MQ, and bisect QMP, QMN by MR, MS respectively, meeting PQ and PQ produced in R and S. Then R is a second point in the locus, since RP: RQ = MP: MQ (278); and S is a third point (280), since SP: SQ=MP: MQ = RP: RQ; so that PQ is divided harmonically in R and S in the given ratio. Also RMS is a right angle (Ex. 113); .. a circumference described upon RS as diameter is the required locus. Hence divide PQ harmonically in R and S, in the given ratio (309), so that SP: SQRP: RQ; etc. 457. (1) If the given lines AB, CD are parallel, draw an intercept C P F K H E P D E B B D EF perpendicular to them, and divide it internally in P, and externally in P', in the given ratio, so that PE : PF = P'E: P'F. The parallels drawn through P and P' to the given lines will be the required locus, which, with the given lines, divides the intercept harmonically. (2) If the given lines intersect in O, draw the intercept EF perpendicular to the bisector of AOC, one of the angles formed, and divide EF at P in the given ratio. A line that passes through P and O is part of the required locus. Draw PH, PK perpendicular to AB, CD respectively; then, by similar triangles, PH: PK=PE : PF. By proceeding similarly with the bisector of C G H B D ZAOD, we can obtain the other part of the locus. 458. Let AB, CD intersect in O. At O draw OE to OC, a side of AOC, one of the angles formed, and make OE equal to the given sum. Through E draw EF || to OG to meet OA in F, and draw FG 1 to the bisector of ZAOC. FG is part of the required locus. From F draw the altitude FH to OG, an arm of ▲ OFG which is isosceles (Ex. 118). FH is equal to OE (136), and the sum of the altitudes from any point P to the arms OA, OC, is equal to FH (Ex. 149), is equal to OE, is equal to the given line. The rest of the locus is found in a similar way. F E C B G E H 459. Let AB, CD intersect in O. At O draw OE perpendicular to OC, a side of AOC, one of the angles formed, and make OE equal to the given difference. Through E draw EF parallel to OC to meet OA in F, and draw FP the bisector of ZAFG. FP is part of the required locus. From P, any point in FP, draw PH, PK perpendicular to OA, OC respectively. Then PH = PG (101), and PK = PG + GK = PH+ OE. The rest of the locus may be found in a similar way. B E 460. In the given ✪ ✪ place a chord AB equal to 4 twice a side of the given square, and find its mid point C. The circumference described from O as center, with radius OC, is the required locus. For this circumference is the locus of the mid points of all chords in OO equal to AB (218). Now if any secant DE be drawn through C, then rect. DC. CE≈rect. AC. CB≈ AC2 a square equal to the given square. EXERCISES, pp. 171-185. = ... EA FD 461. Since AD = BC, and EF BC, EF = AD; (Ax. 2); also FC = BE; .·. rt. ^ AEB = rt. ▲ DFC; etc. (327). 462. Let AC be the rectangle D and EG the par'm. Draw HK, GL 1 to EF; then rect. KG: rect. AC (317), and par'm EG (327); .. par'm EG≈rect. AC. Α C H BE K G F Ꮮ 463. Since AB = AE + EB, rect. AB. (AE - EB) rect. (AE+ EB)(AE — EB) ≈ AE2 – EB2 (337). 464. Since AE = AB – EB, rect. AE (AB + EB)≈rect. (ABEB) (AB+ EB) ≈ AB2 – EB2. D must be equal to 180° 62° 54' 23' = 117° 5' 37". 466. Let M and N denote the two lines; then (M + N)2 − (M − N)2 ≈ M2 + N2 + 2 M · N − (M2 + N2 467. Since A and ADC are equal to 469. Since AD : A'D' = AB : A'B' = m : n, AD2 : A'D12 = m2 : n2. 470. If P: P'm : n, then AC: A'C' = √m : √ñ (346). 471. In the diagram for Art. 349, left-hand figure, suppose BC drawn 1 to AP; then BC= PQ, and CP BQ; .. AC = AP — BQ; ... AB2 – PQ2≈ AB2 – BC2 ≈ AC2 ≈ (AP – BQ)2. 472. AB2 + BC2 ≈ AE2 + EB2 + 2 AE · EB + CE2 + EB2 ≈ AE2 + CE2 + 2 EB2 + 2 AE · EB ≈ AC2 + 2 AB · EB. Draw DF to AB; then EF or CD = AE - AF, and DF = CE; .. CD2 + AD2≈ AE2 + AF2 - 2 AE AF + CE2 + AF2 AC2 −2 AF · (AE — AF)≈ AC2−2 CD(AE— CD). 473. AB2 + BC2 − (CD2 + AD2)≈ AC2 + 2 AB · EB – AC2 + 2 CD · (AE — CD)~2 AB(AB—AE)+2 CD(AE— CD). 474. Draw BP' 1 to AC; then (350) AB2 ≈ AC2 + BC2 – 2 AC. CP≈ BC2 + AC2 . 2 BC. CP; .. AC· CP'≈ BC. CP. (Ax. 1, Ax. 3). 475. If AB2≈ AC2+3 PC2, it is equivalent C to AP2 + 4 PC2, since AC2 = AP2 + PC2, while AB2≈ AP2 + BP2; then BP2=4 PC2; ... BP = 2 PC; .. P .B P trisects BC. 476. In the diagram for Ex. 474, since ▲ APC, BP'C are similar, AP: BP' = CP: CP' = AC : BC. 2 477. Draw CF 1 to AB produced; then AF = AB + BF, and BF = BC, since, CBF being 60° and Fa right angle, CBF is half of an equilateral triangle. AC2≈ AF2 + CF2 ≈ AB2 + BF2 + 2 AB · BF÷CF2. But BF2 + CF2 ≈ BC2, and 2 AB. BF≈ AB · BC; .. AC2 ≈ AB2 + BC2 + AB · BC. 478. Draw CFL to AB; then AF-AB-BF=1}BC. AC2≈ AF2+ CF2≈AB2+BF2 −2 AB. BF÷CF2≈ AB2+BC2-AB. BC, as above. 479. Let BC be the base of an isosceles ▲ ABC; draw CD1 to AB; then BD is the projection of BC on AB, and BC2≈ AB+ AC2 2 AB · AD≈2(AB2 – AB · AD)≈ 2 AB · BD. = 480. Let AC, BD, the diagonals of parallelogram ABCD, intersect in 0. Since AO CO, AB2 + BC2≈2 AO2 + 2 BO2. Similarly AD2 + CD2≈2 AO2 + 2 DO2 or 2 B02; .. AB2+ BC2 Ᏼ + AD2 + CD2≈≈ 4 AO2 + 4 BO2 ≈ = AC2 -2 + BD2. 481. In AABC let AD be the median drawn to BC. A ADB≈ AADC, since they have equal bases, DB, DC (Hyp.), and the same altitude (316). 482. In ▲ ABC, let AD, BE, CF, be the medians drawn to BC, AC, AB, respectively. Now AB2 + AC2 ≈ 2 (AD2 + BD2), AB2 + BC2≈ 2 (AE2+BE2), and AC2+BC2 2(AF2 + CF2); ... 2(AB2+AC2+ BC2)≈2 (AD2 + BE2 + CF2) + 1⁄2 (AB2 + AC2 + BC2) ; ... 3 (AB2 + AC2 + BC2)≈4 (AD2 + BE2 + CF2). 2 483. Since 2 BC2 ≈ 2 (AB2 + AC2 — AB · AE - AC · AD), BC2≈ AB(AB – AE) + AC(AC – AD)≈ AB · BE + AC · CD. 484. If AB > AC> BC, then (AB – AC)2 + (AB − BC)2+(AC — BC)2>0; ... AB2 + AC2 − 2 AB · AC + AB2 + BC2 – 2 AB · BC + AC2+BC2-2 AC. BC>0; 2(AB2+AC2+BC2)>2(AB · AC+ AB · BC + AC · BC) ; ··. AB2 + etc. > AB AC + etc. 485. Let AD be the median to hypotenuse BC of rt. ▲ ABC. Since AD = BC (Ex. 144), AD2 = } BC2. THEOREMS, p. 186. 486. Let the diagonals AC, BD of parallelogram ABCD intersect in O. Since OB = OD, ^ AOB≈▲AOD (332). Similarly ▲AOB≈ A BOC, and ▲ AODA DOC. = 487. In ▲ ABC, DEF let AB, BC = DF, FE respectively, and ZB be supplement of F. Produce EF to G, so that FG BC, and join DG. Since ZDFG is supplement of F, ZDFG = ≤B; ... ▲ DFG = ▲ ABC (66). But since FG = BC = EF, ▲ DFG≈ ▲ DEF (332) ; AABC ADEF. = 488. Let P be the given point in par'm ABCD. Through P draw QPR || to AB. ▲ PAB par'm AR (331), and ▲ PCD≈1⁄2 par'm DR; :. ▲ PAB + ^ PCD≈ }(AR + DR)≈1 par'm ABCD. Similarly ▲ PAD + ▲ PBC≈ } par'm ABCD; etc. 489. Let P be the point in diagonal AC of par'm ABCD, through which EPF, GPH are drawn || to AB, BC respectively. Since (140) ▲ ABC ACD, ▲ PAG ≈ ▲ PAE, and ▲ PCF≈ ▲ PCH, ▲ ABC −(^ PAG + ▲ PCF)≈ ▲ ACD − (^ PAE + ^ PCH) ; .·. par'm PB par'm PD. 490. Let EF join the mid points E, F of the base or parallel sides of trapezoid ABCD. Join FA, FB. Since (332) A FEA≈ Δ ΡΕΒ, and Δ AFD~ Δ BFC, Δ FEA + Δ AFD ~ Δ FEB + ABFC; i.e., AEFD≈ BEFC. 491. Let EF, FG, GH, HE, join the mid points of the sides AB, BC, CD, DA, of the quadrilateral ABCD. Draw the diagonals AC, |