Two Parallels in a Triangle and One More: What Is It About?
A Mathematical Droodle

 

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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

Two Parallels in a Triangle and One More

The applet illustrates the following problem

  Let X be a point on the side BC of a triangle ABC. The parallel through X to AB meets CA at V and the parallel through X to AC meets AB at W. Let D = BV ∩ XW and E = CW ∩ XV. Prove that DE is parallel to BC and

  1 / DE = 1 / BX + 1/ CX.

The problem has been proposed Francisco Javier García Capitán, Spain for Mathematical Reflections (2, 2009).

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Two pairs of intersecting lines (AC, CW) and (CW, BC) are crossed by two parallel lines AB and VX. This gives

  EV / AW = CE / CW = EX / BW,

implying

  EV / EX = AW / BW.

On the other hand, two parallels WX and AC cross AB and BV. This gives

  DV / BD = AW / BW.

By transitivity,

  EV / EX = DV / BD.

It follows that indeed DE||BC.

Let P be the intersection of AB and DE. Since PE||BX and EX||BP, BXEP is parallelogram and EP = BX. Further, as above,

  DP / DE = BX / CX,

or,

  (BX - DE) / DE = BX / CX,

which can be rewritten as

  BX / DE = 1 + BX / CX,

And, finally, dividing by BX, we get

  1 / DE = 1 / CX + 1 / BX,

as required.

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

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