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ContentElementary Geometry
Elementary GeometryMiscellaneous Propositions931 To divide the triangle π΄π΅πΆ in a given ratio by a line ππ drawn parallel to any given line π΄πΈ.Make π΅π· to π·πΆ in the given ratio. Then make π΅π a mean proportional to π΅πΈ and π΅π·, and draw ππ parallel to πΈπ΄. Proof: π΄π· divides π΄π΅πΆ in the given ratio (VI. 1). Now π΄π΅πΈβΆππ΅πβ·π΅πΈβΆπ΅π·VI. 19 or β·π΄π΅πΈβΆπ΄π΅π·; therefore ππ΅π=π΄π΅π· 932 If the interior and exterior vertical angels at π of the triangle π΄ππ΅ be bisected by straight lines which cut the base in πΆ and π·, then the circle circumscribing πΆππ· gives the locus of the vertices of all triangles on the base π΄π΅ whose sides π΄π, π΅π are in a constant ratio. Proof: The β πΆππ·= 12(π΄ππ΅+π΅ππΈ) =a right angle therefore π lies on the circumference of the circle, diameter πΆπ· (III 31). Also π΄πβΆπ΅πβ·π΄πΆβΆπΆπ΅β·π΄π·βΆπ·π΅ (VI 3 and A.) a fixed ratio. 933 π΄π· is divided harmonically in π΅ and πΆ; i.e. π΄π·βΆπ·π΅β·π΄πΆβΆπΆπ΅; or the whole line is to one extreme part as the other extreme part is to the middle part. If we put π, π, π of the lengths π΄π·, π΅π·, πΆπ·, the proportion is expressed algebraically by πβΆπβ·πβπβΆπβπ, which is equivalent to 1π+ 1π= 2π934 Also π΄πβΆπ΅π=ππ΄βΆππΆ=ππΆβΆππ΅ and π΄π2βΆπ΅π2=ππ΄βΆππ΅VI.19 π΄π2βπ΄πΆ2βΆπΆπ2βΆπ΅π2βπ΅πΆ2VI.3 & π΅ 935 If π be the centre of the inscribed circle of the triangle π΄π΅πΆ, and if π΄π produced meet the circumscribed circle, radius π , in πΉ; and if πΉππΊ be a diameter, and π΄π· perpendicular to π΅πΆ; then πΉπΆ=πΉπ=πΉπ΅=2π π΄2i β πΉπ΄π·=πΉπ΄π= π΄2(π΅βπΆ) and β πΆπ΄πΊ= π΄2(π΅+πΆ)ii Proof of [i]: β πΉππΆ=ππΆπ΄+ππ΄πΆ But ππ΄πΆ=ππ΄π΅=π΅πΆπΉIII.21 β΄πΉππΆ=πΉπΆπ; β΄πΉπΆ=πΉπ; Similarly πΉπ΅=πΉπ Also β πΊπΆπΉ is a right angle, and πΉπΊπΆ=πΉπ΄πΆ= 12π΄III.21 β΄πΉπΆ=2π π΄2936 If π , π be the radii of the circumscribed and inscribed circles of the triangle π΄π΅πΆ (see last figure), and π, π the centres; then ππ2=π 2β2π π Proof: Draw ππ» perpendicular to π΄πΆ; then ππ»=π. By the isosceles triangle π΄ππΉ, ππ2=π 2βπ΄πβ ππΉ (922, iii), and ππΉ=πΉπΆ (935,i), and by similar triangles πΊπΉπΆ, π΄ππ», π΄πβΆππ»β·πΊπΉβΆπΉπΆ; therefore π΄πβ πΉπΆ=πΊπΉβ ππ»=2π π Sources and Referenceshttps://archive.org/details/synopsis-of-elementary-results-in-pure-and-applied-mathematics-pdfdriveΒ©sideway ID: 210900018 Last Updated: 9/18/2021 Revision: 0 Ref: References
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