But EF = AG + GB=2AG or 2GB. It is similarly proved that EF = 2AH or 2BH. COR. 4.-The transverse and conjugate axis are bisected in the centre. AC = CB (II. Def. 2), and EA = BF; therefore EC = CF. Also in the triangles ACG, ACH, AG = AH, therefore the angles at G and H are equal, those at C are right angles, and AC is common; hence (Pl. Ge. I. 26) CG= CH. Cor. 5.-A perpendicular to the transverse at one of its extremities is a tangent to the ellipse. For if I be a point in that perpendicular, AI 7 AF, and BI BF; therefore AI +IB AF+ FB or EF. CoR. 6.—The square of half the conjugate axis is equal to the rectangle under the segments into which the transverse is divided in one of the foci. For CGP = AGP — ACP = EC2-AC2 = EA • AF, or EB: BF (Pl. Ge. II. 5, Cor.) Cor. 7.—The distance of the foci is a mean proportional between the sum and difference of the transverse and conjugate axis. For AC2 = AGP GC2 = (AG + GC) (AG - GC) (Pl. Ge. II.5, Cor), or AG +GC: AC = AC: AG - GC; and the doubles of these terms are also proportional. PROPOSITION II. The straight line which bisects the angle adjacent to that which is contained by two straight lines, drawn from any point in the ellipse to the foci, is a tangent to the curve in that point. Let D be any point in the curve, from which AD, DB, are drawn to the foci, and let the angle BDI adjacent to ADB, be bisected by the line DT; then is DT a tangent E to the ellipse in the point D. >O.. For, let any other LA B point R be assumed in DT, and DI being made equal to DB, let BI, RA, RB, and RI, be drawn. The line DNT, which bisects the vertical angle of the isosceles triangle BDI, also bisects the base BI at right angles. Hence RB = RI (Pl. Ge. I. 4), and AR + RB = AR + RI, and therefore greater than AÍ (Pl. Ge. I. 20) or EF (II. 1). Consequently the point R is without the ellipse (II. 1, Cor. 2). Cor. 1.-A perpendicular to the conjugate axis, at one of its extremities, is a tangent to the ellipse. For this line bisects the angle adjacent to that formed by lines drawn to the foci from this point. COR. 2. The method of drawing a tangent from a given point in the curve, also of drawing a tangent parallel to a line given in position, is evident. When the tangent is to be parallel to a given line, draw from the focus B a line BI perpendicular to the line, and make AI =EF; join AI, and D will be the point of contact. Cor. 3.-There cannot be more than one tangent to the ellipse at the same point. For the sum AD + DB of the lines from A and B, to a point D in RN, which make equal angles with it, is less than the sum of any other two lines drawn from A and B to any other point as R in RN. Now, if another tangent can be drawn through D, AD and DB would not make equal angles with it, but some other two lines AD', D'B, to some point D' in it, would make equal angles with it. But AD' + D'B would be less than AD + DB, and hence the point D' would be within the curve (II. 1, Cor. 2), and the line would not be a tangent. Cor. 4.-Every tangent bisects the angle adjacent to that contained by straight lines drawn to the foci from the point of contact, or, which is the same, these lines make equal angles with the tangent. Cor. 5.--A straight line drawn from the centre to meet a tangent, and parallel to the line joining the point of contact, and one of the foci, is equal to half the trans verse axis. For, since BC = CA, and BN=NI, the line that joins CN will be parallel to AD, and equal to the half of AI. Cor. 6. A perpendicular to a tangent, from one of the foci, and a parallel to the line joining the other focus, and the point of contact from the centre, meet the tangent in the same point. Cor. 7.-If a cord pass through one of the foci, and the tangents at its extremities be produced to meet, the straight line that joins the point of concourse and the focus, will be perpendicular to the cord. • Let the tangents at the extremes of the cord DBO, passing through the focus, meet in the point T, and let a line TI be perpendicular to AD; then, because DT, TO, bisect the exterior angles of the triangle ADO, AI is equal to half its perimeter (Pl. Ge. VI. h) = AD + DB. Hence DB= DI, and triangles TDI, TDB, equal in every respect, and angle TBD = TID, or a right angle. PROPOSITION III. From any point in an ellipse, a perpendicular being let fall upon the transverse, and straight lines drawn to the foci, as half the transverse is to the eccentricity, so is the distance of the centre from the perpendicular to half the difference of the straight lines drawn to the foci. Let D be the point in the curve, from which DA, DB, are drawn to the foci, and DK perpendicular to EF, then CF: CB=CK:1 (AD-DB). For, let EL = AD, then LF / = DB, and 2CL = AD ELA CÍ K BF DB. But (AD + DB).(AD - DB) = AB. 2CK (Pl. Ge. II. c), or EF. 2CL = AB · 2CK; therefore EF : AB = 2CK : 2CL, and CF : CB = CK:CL. PROPOSITION IV. Any two cords which pass through one of the foci of an ellipse, are to one another directly as the rectangles under their segments. Let DO, MI, be the two cords passing through the focus B, DO: MI = DB · BO:MB • Bİ.. For, let the tangents at D, 0, meet each other in R, |