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Explanation

The applet is supposed to suggest the following attractive fact (and its proof):

 ABC is an isosceles right angled triangle with the right angle at C. Points D and E, equidistant form C, are chosen arbitrarily on AC and BC. Perpendiculars from D and C to AE meet the hypotenuse AB at K an L. Prove that KL = BL.

### 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?

To see why this is so, extend AC to AF with CF = CD. Triangles ACE and BCF are equal and are obtained from each other with a rotation through 90°. This is then the angle between their corresponding elements. In particular, BF is perpendicular to AE and is therefore parallel DK and CL. This gives two lines AB and AC crossed by three parallel lines, DK, CL, BF. Since CD = CF, so also KL = BL.

### References

1. R. Honsberger, Mathematical Delights, MAA, 2004, pp.89-90
2. V. V. Prasolov, Problems in Planimetry, v 1, Nauka, Moscow, 1986, p. 30 (in Russian)