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Elementary Geometry
 Miscellaneous Propositions
 Sources and References

Elementary Geometry

Miscellaneous Propositions

931 image To divide the triangle 𝐴𝐡𝐢 in a given ratio by a line π‘‹π‘Œ drawn parallel to any given line 𝐴𝐸.
Make 𝐡𝐷 to 𝐷𝐢 in the given ratio. Then make π΅π‘Œ a mean proportional to 𝐡𝐸 and 𝐡𝐷, and draw π‘Œπ‘‹ parallel to 𝐸𝐴.
Proof: 𝐴𝐷 divides 𝐴𝐡𝐢 in the given ratio (VI. 1). Now π΄π΅πΈβˆΆπ‘‹π΅π‘Œβˆ·π΅πΈβˆΆπ΅π·VI. 19 or ∷𝐴𝐡𝐸∢𝐴𝐡𝐷; therefore π‘‹π΅π‘Œ=𝐴𝐡𝐷 932 image If the interior and exterior vertical angels at 𝑃 of the triangle 𝐴𝑃𝐡 be bisected by straight lines which cut the base in 𝐢 and 𝐷, then the circle circumscribing 𝐢𝑃𝐷 gives the locus of the vertices of all triangles on the base 𝐴𝐡 whose sides 𝐴𝑃, 𝐡𝑃 are in a constant ratio.
Proof: The βˆ πΆπ‘ƒπ·=12(𝐴𝑃𝐡+𝐡𝑃𝐸) =a right angle therefore 𝑃 lies on the circumference of the circle, diameter 𝐢𝐷 (III 31). Also π΄π‘ƒβˆΆπ΅π‘ƒβˆ·π΄πΆβˆΆπΆπ΅βˆ·π΄π·βˆΆπ·π΅ (VI 3 and A.) a fixed ratio. 933 𝐴𝐷 is divided harmonically in 𝐡 and 𝐢; i.e. 𝐴𝐷∢𝐷𝐡∷𝐴𝐢∢𝐢𝐡; or the whole line is to one extreme part as the other extreme part is to the middle part. If we put π‘Ž, 𝑏, 𝑐 of the lengths 𝐴𝐷, 𝐡𝐷, 𝐢𝐷, the proportion is expressed algebraically by π‘ŽβˆΆπ‘βˆ·π‘Žβˆ’π‘βˆΆπ‘βˆ’π‘, which is equivalent to 1π‘Ž+1𝑏=2𝑐 934 Also π΄π‘ƒβˆΆπ΅π‘ƒ=π‘‚π΄βˆΆπ‘‚πΆ=π‘‚πΆβˆΆπ‘‚π΅ and 𝐴𝑃2βˆΆπ΅π‘ƒ2=π‘‚π΄βˆΆπ‘‚π΅VI.19 𝐴𝑃2βˆ’π΄πΆ2βˆΆπΆπ‘ƒ2βˆΆπ΅π‘ƒ2βˆ’π΅πΆ2VI.3 & 𝐡 935 image If 𝑄 be the centre of the inscribed circle of the triangle 𝐴𝐡𝐢, and if 𝐴𝑄 produced meet the circumscribed circle, radius 𝑅, in 𝐹; and if 𝐹𝑂𝐺 be a diameter, and 𝐴𝐷 perpendicular to 𝐡𝐢; then 𝐹𝐢=𝐹𝑄=𝐹𝐡=2𝑅sin𝐴2i ∠𝐹𝐴𝐷=𝐹𝐴𝑂=𝐴2(π΅βˆ’πΆ) and ∠𝐢𝐴𝐺=𝐴2(𝐡+𝐢)ii Proof of [i]: βˆ πΉπ‘„πΆ=𝑄𝐢𝐴+𝑄𝐴𝐢 But 𝑄𝐴𝐢=𝑄𝐴𝐡=𝐡𝐢𝐹III.21 βˆ΄πΉπ‘„πΆ=𝐹𝐢𝑄; ∴𝐹𝐢=𝐹𝑄; Similarly 𝐹𝐡=𝐹𝑄 Also ∠𝐺𝐢𝐹 is a right angle, and 𝐹𝐺𝐢=𝐹𝐴𝐢=12𝐴III.21 ∴𝐹𝐢=2𝑅sin𝐴2 936 If 𝑅, π‘Ÿ be the radii of the circumscribed and inscribed circles of the triangle 𝐴𝐡𝐢 (see last figure), and 𝑂, 𝑄 the centres; then 𝑂𝑄2=𝑅2βˆ’2π‘…π‘Ÿ Proof: Draw 𝑄𝐻 perpendicular to 𝐴𝐢; then 𝑄𝐻=π‘Ÿ. By the isosceles triangle 𝐴𝑂𝐹, 𝑂𝑄2=𝑅2βˆ’π΄π‘„β‹…π‘„πΉ (922, iii), and 𝑄𝐹=𝐹𝐢 (935,i), and by similar triangles 𝐺𝐹𝐢, 𝐴𝑄𝐻, π΄π‘„βˆΆπ‘„π»βˆ·πΊπΉβˆΆπΉπΆ; therefore 𝐴𝑄⋅𝐹𝐢=𝐺𝐹⋅𝑄𝐻=2π‘…π‘Ÿ

Sources and References

https://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdrive

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ID: 210900018 Last Updated: 9/18/2021 Revision: 0 Ref:

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References

  1. Hilbert, D. (translated by Townsend E.J.), 1902, The Foundations of Geometry
  2. Moore, E.H., 1902, On the projective axioms of geometry
  3. Fitzpatrick R. (translated), Heiberg J.L. (Greek Text), Euclid (Author), 2008, Euclid's Elements of Geometry
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