from the greater segment of the base, and arc, parts be cut off respectively equal to the less; the remaining part of the base shall be equal to the chord of the remaining arc. From any point B in the arc ABC, let BD be drawn perpendicular to AC; make BF BC, and DE=DC; join AF; AF = will be equal to AE. EDC Join FE, EB, FB, BC. Since the arc BC= the arc BF, the straight line BC= BF; and DE being equal to DC, and DB common, and at right angles to EC, .. BE = BC= BF, and the angle BFE is equal to the angle BEF. Now since AFBC is a quadrilateral figure inscribed in a circle, the angles AFB, ACB are equal to two right angles, and.. equal to AEB, CEB, of which ACB = CEB, .. AFB = AEB; but BFE = BEF, consequently AFE=AEF; whence AF-AE. :. (72.) If from the point of bisection of any arc of a circle a perpendicular be drawn to the diameter, which passes through one extremity; it will bisect the segment of the chord cut off by the line joining the point of bisection of the arc and the other extremity of the diameter. Let AC be the arc bisected in D. Join AC; and from D draw DE perpendicular to the diameter AB, and meeting AC in G; B join BD; AG=GF. D Because AC is bisected in D, the angle CAD is equal to the angle DBA, i. e. to the angle EDA (Eucl. vi. 8.),.. the right-angled triangle ADF is equiangular to the two triangles BED, DEA, .. the angle GFD= GDF, and consequently GD=GF; also GAD=GDA, .. AG = GD, whence AG = GF. (73.) In a given circle to draw a chord parallel to a straight line given in position; so that the chord and perpendicular drawn to it from the centre may together be equal to a given line. Let O be the centre of the circle, OA the straight line given in position; draw OB perpendicular to it, and equal to the given line. Take OA equal to the half of OB; and join AB, cutting the circle in C; through C draw CD parallel to OA; CD is the chord required. E B Because OA is half of OB, and OA, EC are parallel, .. (Eucl. vi. 2.) EC is half of EB, and DC = EB; therefore DC and OE together are equal to BE and OE together, i. e. to BO, or to the given line. (74.) Through a given point within a given circle, to draw a straight line such that the parts of it intercepted between that point and the circumference may have a given ratio. Let P be the given point within the circle ABD. Through P draw the diameter APB, and take AP: PC in the given ratio. With P as centre, and radius equal to a mean proportional between BP and PC, P A B 同 describe a circle cutting ADB in D; join DP, and produce it to E; DE is the chord required. Since BP PD :: PD: PC, and (Eucl. iii. 35.) BP : PD :: PE : PA, PA :: PD: PC, the given ratio. .. PE and alt. PE: PD :: AP: PC, i. e. in COR. Since one circle cuts another in two points, there will be two chords which answer the conditions. If C coincides with A, the ratio is one of equality, and DE will be perpendicular to AB. (75.) From two given points in the circumference of a given circle, to draw two lines to a point in the circumference, which shall cut a line given in position, so that the part of it intercepted by them may be equal to a given line. Let A, B be the given points in the circumference of the circle ABC; DE the line given in position. From B draw BF parallel to DE, and equal to the given line. Join AF; and on D it describe a segment of a circle AGF containing an angle equal to the angle in the segment ACB; and let it cut DE in G. Join AG, and produce it to C; and join BC cutting DE in H. AC, BC are the lines required. Join GF. Since the angle AGF=ACB, GF is parallel to CB; but FB is parallel to GH, whence FGHB is a parallelogram, and GH=FB. (76.) If a chord and diameter of a circle intersect each other at any angle, and a perpendicular to the chord be drawn from either extremity of it, meeting the circumference and diameter produced; the whole perpendicular has to the part of it without the circle, the same ratio that the greater segment of the chord has to the less. Let the diameter AB and chord DE intersect each other at C; and from D draw DG perpendicular to DE, then GD GF :: DC: CE. Through F draw FI parallel to DE, and meeting the diameter in I. Join FE, cutting the diameter in O. Since the angle FDE is a right angle, FE is a diameter and O is the centre. And since the angle IFO is equal to the alternate angle OEC, and the angles at O are equal, and FOOE, .. the triangles OFI, OEC are equal, and CE= FI. And since FI is parallel to DC, (Eucl. vi. 2.) GD: GF :: DC : (FI=) CE. (77.) If from the extremities of any chord of a circle, perpendiculars to it be drawn and produced to cut a diameter; and from the points of intersection with the diameter lines be drawn to a point in the chord, so as to make equal angles with it; these lines together will be equal to the diameter of the circle. B Let AB be any chord of the circle ABC; draw AE and BF perpendicular to it, meeting the diameter CD in E and F; from which let the lines EH, FH be drawn making equal angles with AB; EH and HF together are equal to CD. Take O the centre, and join BO, and produce it; it will meet AЕ produced in G. Produce EH, FB to meet in I. Then since the angle IHB=AHE=FHB, and HB is perpendicular to FI, the triangles FHB, HBI are equal, and FH= HI. And since EG is parallel to FB, the angle EGO OBF, and the vertical = angles at O are equal, and GOOB, .. EG=FB=BI; whence EI= GB, and .. EH, HF together are equal to EI i. e. to GB or CD the diameter of the circle. (78.) If from a point without a circle two straight lines be drawn, one of which touches and the other cuts the circle; a line drawn from the same point in any direction, equal to the tangent, will be parallel to the chord of the arc intercepted by two lines drawn from its other extremity to the former intersections of the circle. From the point A let AB, AD be drawn, of which AB touches the circle BCD, and AD cuts it; and draw AE= B AB, in any direction; join CE, DE, cutting the circle in F and G; the chord FG will be parallel to AE. K D G |