60^o Breeds 90^o: What is this about?
A Mathematical Droodle

Explanation

The applet is supposed to suggest the following problem:

(1)	Angle bisectors AF, BG, and CH in ΔABC are drawn. It is known that A = 120o. Prove that angle GFH is right.

Extend AB beyond A to, say, X. Since BAC = 120^o, CAX = 60^o. Also CAF = 60^o, so that AC is the external angle bisector at A of ΔABF. The external angle bisector at A intersects the (interior) angle bisector at B in an excenter of ΔABF. It follows that G is an excenter of that triangle. Another external angle bisector passes through G: that of the exterior angle at F. Therefore, FG is the bisector of AFC. Similarly, FH is the bisector of AFB. Since the two angles are supplementary, so that the angle between FG and FH is indeed 90^o.

References

1. D. Fomin, A. Kirichenko, Leningrad Mathematical Olympiads 1987-1991, MathPro Press, 1994, pp. 38, 146-7