Two Parallels in a Triangle and One More: What Is It About?
A Mathematical Droodle
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Explanation
Copyright © 1996-2009 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
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 http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.
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
On the other hand, two parallels WX and AC cross AB and BV. This gives
By transitivity ,
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,
or,
(BX - DE) / DE = BX / CX,
which can be rewritten as
And, finally, dividing by BX, we get
1 / DE = 1 / CX + 1 / BX,
as required.
Copyright © 1996-2009 Alexander Bogomolny
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