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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 ๐ด๐ธโ ๐ด๐น=๐ด๐ต2III.36 ๐ด๐ต2=๐ต๐ทโ ๐ท๐ถ+๐ด๐ท2923 and ๐ต๐ทโ ๐ท๐ถ=๐ธ๐ทโ ๐ท๐นIII.35 949 If ๐ผ, ๐ฝ, ๐พ, in the same figure, be the perpendiculars to the sides of ๐ด๐ต๐ถ from any point ๐ธ on the circumference of the circle, then ๐ฝ๐พ=๐ผ2ProofDraw the diameter ๐ต๐ป=๐; then ๐ธ๐ต2=๐ฝ๐, because ๐ต๐ธ๐ป is a right angle. Similarly ๐ธ๐ถ2=๐พ๐. But ๐ธ๐ตโ ๐ธ๐ถ=๐ผ๐ (VI. D), therefore โฏ 950Cor: If ๐ด๐ถ=๐โ ๐ด๐ธ, then ๐ต๐ธ=(๐+1)๐๐ธ 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 ๐ต๐ธ, ๐ถ๐น perpendicular to the sides, and let them intersect in ๐. Let ๐ด๐ meet ๐ต๐ถ in ๐ท. Circles will circumscribe ๐ด๐ธ๐๐น and ๐ต๐น๐ธ๐ถ, by (III.31); therefore โ ๐น๐ด๐=๐น๐ธ๐ต=๐น๐ถ๐ตIII.21 therefore โ ๐ต๐ท๐ด=๐ต๐น๐ถ=a right angle; i.e. ๐ด๐ is perpendicular to ๐ต๐ถ, and therefore the perpendicular from ๐ด on ๐ต๐ถ passes through ๐. ๐ is called the orthocentre of the triangle ๐ด๐ต๐ถ. CorThe perpendiculars on the sides bisect the angles of the triangle ๐ท๐ธ๐น, and the point ๐ is therefore the centre of the inscribed circle of that triangle.Proof. From (III.21), and the circles circumscribing ๐๐ธ๐ด๐น and ๐๐ธ๐ถ๐ท Sources and Referenceshttps://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdriveยฉsideway ID: 210900020 Last Updated: 9/20/2021 Revision: 0 Ref: References
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