segment of the diameter intercepted between that perpendicular and chord is to the intercepted part of the tangent, as the chord is to the diameter. Let the chord CD be perpendicular to the diameter AB, and let CE touch the circle at C; from any point E in which let EF be drawn perpendicular to AB; FG: CE CD : AB. E B F D H Draw the diameter CH; join HD and draw CI perpendicular to EF. Since EC touches the circle, the angle ECH (Eucl. iii. 18.) is a right angle and .. equal to ICD; whence, taking away from each ICH, the angle ECI=HCD, and EIC, HDC are right angles, .. the triangles ECI, HDC are equiangular, whence IC CE :: DC: CH, : (69.) If a straight line be placed in a circle, and from its extremities perpendiculars be let fall upon any diameter; these perpendiculars together will have to the part of the diameter intercepted between them, the same ratio that a line placed in the circle perpendicular to the former line, has to the former line itself. Let the line CD be placed in the circle ABC, and from its extremities let CE, DF be drawn perpendicular to a diameter AB. From D let DG be drawn perpendicular to DC; then will G CE+DF: EF :: GD: DC. and produce CE to I: join DI, and draw DH perpendicular to CE. Since CI is perpendicular to AB, CE= EI, but HE = DF, .. HI = CE + DF. Now (Eucl. iii. 21.) the angle at G is equal to the angle at I, and CDG, DHI are right angles, .. the triangles CGD, HID are equiangular, and HI: HD: DG: DC,. or CE+DF: EF :: DG : DC. (70.) In a circle to place a straight line of given length, so that perpendiculars drawn to it from two given points in the circumference may have a given ratio. H B F Let A and B be the given points in the circumference of the circle whose centre is 0. Join BA and produce it, and take AC : CB in the given ratio. In the circle place a straight line equal to the given straight line, and from the centre O let fall a perpendicular upon it. With O as centre, and distance equal to this perpendicular describe a circle DG, and from C draw CEDF a tangent to it; then HF is the line required. For (Eucl. iii. 14.) it is equal to the given straight line. And if from A and B, AE, BI be drawn perpendicular to CF, they are parallel to each other, and the triangles CAE, CBI are similar, :. AE : BI :: CA: CB, i. e. in the given ratio. (71.) If from any point in the arc of a segment of a circle a line be drawn perpendicular to the base; and 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; join BD; AG = GF. 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, 04 the straight line given in position; draw OB perpendicular to it, and equal to the given line. Take OA equal half of OB, and join AB cutting the circle in C; through C draw CD parallel to OA; CD is the chord required. ΟΙ D E 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 D C P E B 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, the given ratio. 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. |