. The principles of projective geometry applied to the straight line and conic . Non-2)rojective proofs 353 Conversely, if the above relation holds the point P is on the line joining P^ to P,^.Note. la all cases when an area passes through a zero value its sign must bechanged. {h) If DyCu B-fJ-i, B-sC-i, ... and D^Fi, , B^F-i, ... be any finite straight linesof limited length in a plane, the loetis of a point such that 2PBiC\ = 2PDiFi is astraight line. This result may be easily deduced from (a). 8. At any internal point on the connector of the middle points of the diagonals ofa quadrilat
. The principles of projective geometry applied to the straight line and conic . Non-2)rojective proofs 353 Conversely, if the above relation holds the point P is on the line joining P^ to P,^.Note. la all cases when an area passes through a zero value its sign must bechanged. {h) If DyCu B-fJ-i, B-sC-i, ... and D^Fi, , B^F-i, ... be any finite straight linesof limited length in a plane, the loetis of a point such that 2PBiC\ = 2PDiFi is astraight line. This result may be easily deduced from (a). 8. At any internal point on the connector of the middle points of the diagonals ofa quadrilateral the sum of the areas subtended by one pair of opposite sides is equalto the sum of the areas subtended by the other Let A, L, M be the middle points of the diagonals BD, ^^0 and EF of thequadrilateral. Take P a point on KML inside the quadrilateral. Then, denoting areas of triangles by their three vertices, PBC+ PKC= KI)C+ PBK, and PAB + PKB=ABK+PKA. But PKC=PKA, since L is the middle point of AC; and PKB = PJJK „ K „ „ „ „ BD. •. PDC + PA B = KDC+ KA B. Similarly PA D + PBC=KAD + KBC. But AM U= HAD and KDC= KBC, since K is the middle point of BD. Therefore PDC+ PAB = PAD + PBC. 23 354 Principles of Projective Geometi-y The Circle. 9. (0 If ^^ /«A of opposite sides ofa hii,mgon viscribed in a circle are parallel,then the remaining pair of opposite sidesare also parallel. In the hexagon ABCDEF, let the pairsof sides AB, ED and BC, EF be AD. Since AB and BC are parallelto ED and FE, the angles ABC and FEDare equal. But the angles ADC and ABCtogether equal two right angles as do theangles FAD and FED. Therefore FADand CDA are equal and the lines CD andAF are parallel. (6)
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