(80.) If any chord in a circle be bisected by another, and produced to meet the tangents drawn from the extremities of the bisecting line; the parts intercepted between the tangents and the circumferences are equal. Let AB be bisected in E by CD; and to C and D let tangents be drawn, meeting AB produced in F and G; AF is equal to BG. Find the centre; join OC, OD, F E OE, OF, OG. Since OE is drawn from the centre to the point of bisection of AB (Eucl. iii. 3.) the angle OEF is a right angle; and the angle OCF is a right angle (Eucl. iii. 18.); .. a circle may be described about OEFC. Also since ODG and OEG are right angles. a circle may be described about OEDG; and the angle DOG is equal to the angle DEG in the same segment; but DEG is equal to FEC, i. e. to FOC, .. DOG = FOC; and ODG, OCF are equal being right angles, and OC OD, .. OF=OG, and consequently FE= EG. But AE= EB, .. FA = BG. (81.) If one chord in a circle bisect another, and tangents drawn from the extremities of each be produced to meet; the line joining their points of intersection will be parallel to the bisected chord. Let AB be bisected by the line CD in E, and let the tangents AF, BF meet each other in F, and DG, CG in G. Join GF; GF is parallel to AB. Join AO, CO, GO, FO; then F D B GO bisects CD in H, and OHE is a right angle; for the same reason FO passes through E, and AEO is a right angle. And since FAO is a right angle (Eucl. iii. 18.), and from A, AE is drawn perpendicular to the base, (Eucl. vi. 8. Cor.) FO: OA :: OA : OE, for the same reason, (OC=) OA: OG :: OH : (OC=)OA, .. ex æquo per. FO OG :: OH: OE, : ... the sides of the triangles FOG, OHE about the common angle O are proportional, and consequently the triangles are equiangular, and the angle GFO equal to EHO, and .. a right angle, and equal to the alternate angle FEB,.. AB is parallel to GF. (82.) If from a point without a circle two lines be drawn touching the circle, and from the extremities of any diameter lines be drawn to the points of contact, cutting each other within the circle; the line produced, which joins their intersection and the point without the circle, will be perpendicular to the diameter. From the point P without the circle ABC let there be drawn two tangents PC, PD. From A and B the extremities of a diameter, draw AD, BC to the points of contact, intersecting each other B in E; join PE, and produce it to F; PF is perpendicular to AB. Take O the centre; join CO, DO, CD, DB. Since CPD, COD, are together equal to two right angles, CPD is equal to AOC, BOD together, i. e. to twice ADC, BCD together, .. CPD is equal to the angle at the centre of a circle passing through C, E, and D; and since PC PD, P is the centre itself;.. PE=PD, and the angle PED is equal to the angle PDE. But the angle DBA=PDE=PED=AEF, and the angle at A is common, .. AFE=ADB, and (Eucl. iii. 31.) is ... a right angle. (83.) If on opposite sides of the same extremity of the diameter of a circle equal arcs be taken, and from the extremities of these arcs lines be drawn to any point in the circumference, one of which cuts the diameter, and the other the diameter produced; the distances of the points of intersection from the extremities of the diameter are proportional to each other. On opposite sides of the point A in AB the diameter of the circle ABC let equal arcs AC, AD be taken; from C and D let CE, DE be drawn to any point E in the circumference, of which CE cuts AB produced in F, and DE cuts AB in G; then will AF: FB :: AG : GB. G B H Join AE, BE, and through B draw HBI parallel to AE. Since AEB is a right angle, CEA and BEI are together equal to a right angle, and .. equal to AED, DEB; and since AC= AD, CEA= AED, .. BEI= BED. Again since AE is parallel to IH, the angle EIB is equal to CEA = AED= the alternate angle EHB, .. the two triangles EIB, EHB having two angles in each equal, and one side EB common are equal, and BI=BH. And from the similar triangles AGE, BGH, AG: AE: BG: (BH=) BI, alt. AG: GB :: AE : BI :: AF : BF, since the triangles AEF, FIB are similar. (84.) If from the extremities of any chord in a circle, perpendiculars be drawn to a diameter, and from either extremity of that diameter a perpendicular be drawn to the chord; it will divide it into segments, which are respectively mean proportionals between the segments of the diameter made by the perpendiculars. Let AB be any chord, and CD a diameter of the circle ABC; AE, BF perpendiculars from A and B to the diameter, and DG perpendicular from D to AB; the segment GB is a mean proportional between DF and CE; and AG a mean proportional between DE and CF. Join AC, AD, BC, BD. Then the angle DBA being equal to DCA and the angles DGB, DAC, AEC being right angles, the triangles DGB, DAC, ACE are similar; also the triangles DBF, DBC are similar; whence DF: DB::DB: DC: BG: AC but DB: BG:: AC: CE, .. ex æquo DF : BG :: BG : CE. And in the same manner it may be proved that (85.) If from any point in the diameter of a semicircle, a perpendicular be drawn, meeting the circumference, and on it as a diameter a circle be described, to the centre of which a line is drawn from the farther extremity of the diameter of the semicircle, cutting its circumference; and through the point of intersection another line be drawn from the extremity of the perpendicular, meeting the diameter of the semicircle; this diameter will be divided into three segments which are in continued proportion. From any point D in the diameter AC of the semicircle ABC, let a perpendicular DB be drawn, on which describe a circle DBG. Find its cen H tre H, join HC cutting the circumference in G; join BG and produce it to E; AD: DE :: DE: EC. Join BC, and draw EF parallel to BD; join DG, GF, AB. Since EF is parallel to DB, and DB is bisected in H, .. EF is bisected in 1. Also the angle HBG is equal to the alternate angle GEI and BHG= GIE, .. the triangles BHG, GIE are equiangular, and BH: HG :: EI : IG, or HD: HG :: FI : IG, i. e. the sides about the equal angles are proportional, .. the triangle FGI is equiangular to HDG, and the angle FGI is equal to HGD; whence HI being a straight line, FGD is also. Again the angle GDE is equal to the angle in the alternate segment DBG, whence the triangles BDE, FDE are similar, .. BD : DE :: DE : EF, but AD: DB:: EF: EC, since the tri |