Segment Trisection Induced by Parallels to Medians

What is this about?
A Mathematical Droodle

14 October 2015, Created with GeoGebra

Explanation

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Copyright © 1996-2017 Alexander Bogomolny

The applet attempts to suggest the following problem [Prasolov, p. 13]:

In ΔABC, AK and CL are two medians, P a point on AC, PE||AK and PF||CL (E on BC, F on AB). Prove that EF is divided into three equal parts by the points M and N of intersection with CL and AK.

14 October 2015, Created with GeoGebra

Let Q and R be the intersections of AK with PF and CL with PE and let O be the centroid of ΔABC. AO:KO = 2:1. Since PE||AK triangles CER and CKO are similar, as are triangles CRP and COA, implying the proportions

ER/KO = CR/CO = PR/AO,

so that

PR/ER = AO/KO = 2/1.

Since PF||CL, triangles ENR and EFP are similar, implying

2/1 = PR/ER = FN/NE,

so that NE = EF/3. Similarly, FM = EF/3 and then also MN = EF/3.

References

  1. V. V. Prasolov, Problems in Planimetry I, Nauka, 1986 (in Russian)

Related material
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  • The Nature of Pi
  • Parallelogram and Similar Triangles
  • Two Triples of Similar Triangles
  • Three Similar Triangles
  • Similar Triangles on Sides and Diagonals of a Quadrilateral
  • Point on Bisector in Right Angle
  • |Activities| |Contact| |Front page| |Contents| |Geometry|

    Copyright © 1996-2017 Alexander Bogomolny

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