Two Parallels in a Triangle and One More: What Is It About?
A Mathematical Droodle
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Explanation
Copyright © 1996-2010 Alexander Bogomolny
Two Parallels in a Triangle and One More
The applet illustrates the following problem
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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).
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Two pairs of intersecting lines (AC, CW) and (CW, BC) are crossed by two parallel lines AB and VX. This gives
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EV / AW = CE / CW = EX / BW,
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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,
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(BX - DE) / DE = BX / CX,
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which can be rewritten as
And, finally, dividing by BX, we get
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1 / DE = 1 / CX + 1 / BX,
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as required.
Copyright © 1996-2010 Alexander Bogomolny
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