A first course in projective geometry . cle to any point of therange is perpendicular to the polar of that point. Also all the polars pass through a common point, viz. thepole of the line on which the range is situated. Therefore the polars form a pencil, and since the anglebetween any two straight lines is equal to the angle between POLES AND POLARS 67 perpendiculars to them, the pencil formed by the polars andthat formed by the joins of the centre to the points of therange (shown by the full drawn lines in the figure) are super-posable. But the latter pencil is harmonic; .. so is theother. C
A first course in projective geometry . cle to any point of therange is perpendicular to the polar of that point. Also all the polars pass through a common point, viz. thepole of the line on which the range is situated. Therefore the polars form a pencil, and since the anglebetween any two straight lines is equal to the angle between POLES AND POLARS 67 perpendiculars to them, the pencil formed by the polars andthat formed by the joins of the centre to the points of therange (shown by the full drawn lines in the figure) are super-posable. But the latter pencil is harmonic; .. so is theother. Cor. The poles of the rays of a harmonic pencil forma harmonic range. The theorem of this article is a particular case of a moregeneral one. (See Chap. XIV. § 1.) § 9. The fundamental Harmonic Property of Poleand Polar. Any straight line drawn through a point P to cut a givencircle is divided harmonically by P, the circle, and the polarof P. First Proof. Let KK, the polar of P, cut the chord PRR atQ, and let CP cut KK at V (Fig. 33a).. Then (§ 2) P and V are inverse points with respect to thecircle. Therefore any circle through them cuts the gi\en circleorthogonally. (Chap. V. § 3.) Consider the circle on PQ as diameter. Since it passes through V it cuts the circle KRK ortho-gonally. Let O be the middle point of PQ. 68 PROJECTIVE GEOMETRY Then it follows that ^sq. on tangent from O to the circle KRK = OQ-. .*. PRQR is a harmonic range. (Chap. IV. § 4 (2) Cor.)This theorem is of great importance, and a second proof by Euclidean methods is appended. Second Proof (Caseys). The construction being as before,join CR, CR (Fig. 33a). Then PR. PR = PK^ = PV. PC, since PKC is a right angle. .*. RVCR is a cyclic quadrilateral. .. angle RVP = angle CRR = angle CRR (since CR = CR) = angle CVR (in same segment). .. angle KVR = angle KVR. And since KVP is a right angle, the range PRQR is harmonic. The proof holds equally well when P is inside the circle;the student should draw the fi
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