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`Elementary GeometryβMiscellaneous PropositionsβSources and References`

# Elementary Geometry

## Miscellaneous Propositions

931 To divide the triangle π΄π΅πΆ in a given ratio by a line ππ drawn parallel to any given line π΄πΈ.
Make π΅π· to π·πΆ in the given ratio. Then make π΅π a mean proportional to π΅πΈ and π΅π·, and draw ππ parallel to πΈπ΄.
Proof: π΄π· divides π΄π΅πΆ in the given ratio (VI. 1). Now π΄π΅πΈβΆππ΅πβ·π΅πΈβΆπ΅π·VI. 19 or β·π΄π΅πΈβΆπ΄π΅π·; therefore ππ΅π=π΄π΅π· 932 If the interior and exterior vertical angels at π of the triangle π΄ππ΅ be bisected by straight lines which cut the base in πΆ and π·, then the circle circumscribing πΆππ· 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 πΆππ·=12(π΄ππ΅+π΅ππΈ) =a right angle therefore π lies on the circumference of the circle, diameter πΆπ· (III 31). Also π΄πβΆπ΅πβ·π΄πΆβΆπΆπ΅β·π΄π·βΆπ·π΅ (VI 3 and A.) a fixed ratio. 933 π΄π· is divided harmonically in π΅ and πΆ; i.e. π΄π·βΆπ·π΅β·π΄πΆβΆπΆπ΅; 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·, πΆπ·, the proportion is expressed algebraically by πβΆπβ·πβπβΆπβπ, which is equivalent to 1π+1π=2π 934 Also π΄πβΆπ΅π=ππ΄βΆππΆ=ππΆβΆππ΅ and π΄π2βΆπ΅π2=ππ΄βΆππ΅VI.19 π΄π2βπ΄πΆ2βΆπΆπ2βΆπ΅π2βπ΅πΆ2VI.3 & π΅ 935 If π be the centre of the inscribed circle of the triangle π΄π΅πΆ, and if π΄π produced meet the circumscribed circle, radius π, in πΉ; and if πΉππΊ be a diameter, and π΄π· perpendicular to π΅πΆ; then πΉπΆ=πΉπ=πΉπ΅=2πsinπ΄2i β πΉπ΄π·=πΉπ΄π=π΄2(π΅βπΆ) and β πΆπ΄πΊ=π΄2(π΅+πΆ)ii Proof of [i]: β πΉππΆ=ππΆπ΄+ππ΄πΆ But ππ΄πΆ=ππ΄π΅=π΅πΆπΉIII.21 β΄πΉππΆ=πΉπΆπ; β΄πΉπΆ=πΉπ; Similarly πΉπ΅=πΉπ Also β πΊπΆπΉ is a right angle, and πΉπΊπΆ=πΉπ΄πΆ=12π΄III.21 β΄πΉπΆ=2πsinπ΄2 936 If π, π be the radii of the circumscribed and inscribed circles of the triangle π΄π΅πΆ (see last figure), and π, π the centres; then ππ2=π2β2ππ Proof: Draw ππ» perpendicular to π΄πΆ; then ππ»=π. By the isosceles triangle π΄ππΉ, ππ2=π2βπ΄πβππΉ (922, iii), and ππΉ=πΉπΆ (935,i), and by similar triangles πΊπΉπΆ, π΄ππ», π΄πβΆππ»β·πΊπΉβΆπΉπΆ; therefore π΄πβπΉπΆ=πΊπΉβππ»=2ππ

## Sources and References

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

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

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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