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Explanation

The applet attempts to suggest the following problem:

 Consider the right isosceles triangle ABC and the points M, N on the hypothenuse BC in the order B, M, N, C such that BM² + NC² = MN². Prove that ∠MAN = 45°.

### 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.

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The configuration appears in a problem introduced in the National Olympiad 2001, in Romania, by Mircea Fianu.

Let's rotate ΔABC 90° counterclockwise around A. Since, by construction, ∠BAC = 90° and AB = AC, ΔABC will map onto ΔACC', where CC' is equal and perpendicular to BC. M will map on M' on CC' such that ∠MAM' = 90°.

ΔNCM' being right, we have

 (NM')² = CN² + (CM')² = CN² + BM² = NM².

Therefore, NM' = NM and, by SSS, ΔAMN = ΔAM'N. So that angles MAN and NAM' are equal and both are equal 45°.

(This problem admits a nice generalization.)

### References

1. W. G. Boskoff and B. D. Suceava, A Projectivity Characterized by the Pythagorean Relation, Forum Geometricorum, Volume 6 (2006) 187–190.