output.to from Sideway
Draft for Information Only

# Content

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

# Elementary Geometry

## Miscellaneous Propositions

926 In a given line π΄πΆ, to find a point π whose distance from a point π shall have a given ratio to its distance in a given direction from a line π΄π΅.
Through π draw π΅ππΆ parallel to the given direction. Produce π΄π, and make πΆπΈ in the given ratio to πΆπ΅. Draw ππ parallel to πΈπΆ, and ππ to πΆπ΅. There are two solutions when πΆπΈ cuts π΄π in two points.  Proof. By VI. 2 927 To find a point π in π΄πΆ, whose distance ππ from π΄π΅ parallel to π΅πΆ shall have a given ratio to its distance ππ from π΅πΆ parallel to π΄π·.
Draw π΄πΈ parallel to π΅πΆ, and having to π΄π· the given ratio. Join π΅πΈ cutting π΄πΆ in π, the point required. Proof. By VI. 2 928 To find a point π on any line, straight or curved, whose distances ππ, ππ, in given directions from two given lines π΄π, π΄π΅, shall be in a given ratio.
Take π any point in the first line. Draw ππ΅ parallel to the direction of ππ, and π΅πΆ parallel to that of ππ, making ππ΅ have to π΅πΆ the given ratio. Join ππΆ, cutting π΄π΅ in π·. Draw π·πΈ parallel to πΆπ΅. Then π΄πΈ produced cuts the lines in π, the point required, and is the locus of such points. Proof. By VI. 2 929 To draw a line ππ through a given point π so that the segments ππ, ππ, intercepted by a given circle, shall be in a given ratio.
Divide the radius of the circle in that ratio, and, with the parts for sides, construct a triangle ππ·πΆ upon ππΆ as base. Produce πΆπ· to cut the circle in π. Draw πππ and πΆπ.
Then ππ·+π·πΆ=radius therefore ππ·=π·π But πΆπ=πΆπ therefore ππ· is parallel to πΆπ (I 5, 28) therefore β― by (VI. 2) 930 From a given point π in the side of a triangle, to draw a line ππ which shall divide the area of the triangle in a given ratio.
Divide π΅πΆ in π· in the given ratio, and draw π΄π parallel to ππ·. ππ will be the line required.
π΄π΅π·: π΄π·πΆ= the given ratio (VI. 1), and π΄ππ·=πππ· (I.37); therefore β―

## Sources and References

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

ID: 210900016 Last Updated: 9/16/2021 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

Nu Html Checker 53 na na

Home 5

Management

HBR 3

Information

Recreation

Culture

Chinese 1097

English 337

Computer

Hardware 156

Software

Application 207

Latex 35

Manim 203

Numeric 19

Programming

Web 285

Unicode 504

HTML 65

CSS 65

SVG 9

ASP.NET 270

OS 422

Python 66

Knowledge

Mathematics

Algebra 84

Geometry 32

Calculus 67

Engineering

Mechanical

Rigid Bodies

Statics 92

Dynamics 37

Control

Natural Sciences

Electric 27