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?

Explanation

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.