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ContentElementary Geometry
Elementary GeometryMiscellaneous Propositions947DefinitionA centre of similarity of two plane curves is a point such that any straight line being drawn through it to cut the curves, the segments of the line intercepted between the point and the curves are in a constant ratio. 948Prooffrom 𝐴𝐸⋅𝐴𝐹=𝐴𝐵2III.36 𝐴𝐵2=𝐵𝐷⋅𝐷𝐶+𝐴𝐷2923 and 𝐵𝐷⋅𝐷𝐶=𝐸𝐷⋅𝐷𝐹III.35 949 If 𝛼, 𝛽, 𝛾, in the same figure, be the perpendiculars to the sides of 𝐴𝐵𝐶 from any point 𝐸 on the circumference of the circle, then 𝛽𝛾=𝛼2ProofDraw the diameter 𝐵𝐻=𝑑; then 𝐸𝐵2=𝛽𝑑, because 𝐵𝐸𝐻 is a right angle. Similarly 𝐸𝐶2=𝛾𝑑. But 𝐸𝐵⋅𝐸𝐶=𝛼𝑑 (VI. D), therefore ⋯ 950Cor: If 𝐴𝐶=𝑛⋅𝐴𝐸, then 𝐵𝐸=(𝑛+1)𝑂𝐸 951 For, by the last theorem, any one of these lines is divided by each of the others in the ratio of two to one, measuring from the same extremity, and must therefore be intersected by them in the same point. This point will be referred to as the centroid of tje triangle. 952 Draw 𝐵𝐸, 𝐶𝐹 perpendicular to the sides, and let them intersect in 𝑂. Let 𝐴𝑂 meet 𝐵𝐶 in 𝐷. Circles will circumscribe 𝐴𝐸𝑂𝐹 and 𝐵𝐹𝐸𝐶, by (III.31); therefore ∠𝐹𝐴𝑂=𝐹𝐸𝐵=𝐹𝐶𝐵III.21 therefore ∠𝐵𝐷𝐴=𝐵𝐹𝐶=a right angle; i.e. 𝐴𝑂 is perpendicular to 𝐵𝐶, and therefore the perpendicular from 𝐴 on 𝐵𝐶 passes through 𝑂. 𝑂 is called the orthocentre of the triangle 𝐴𝐵𝐶. CorThe perpendiculars on the sides bisect the angles of the triangle 𝐷𝐸𝐹, and the point 𝑂 is therefore the centre of the inscribed circle of that triangle.Proof. From (III.21), and the circles circumscribing 𝑂𝐸𝐴𝐹 and 𝑂𝐸𝐶𝐷 Sources and Referenceshttps://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive©sideway ID: 210900020 Last Updated: 9/20/2021 Revision: 0 Ref: ![]() References
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