Rectangle in Arbelos: What is this about?
A Mathematical Droodle
What if applet does not run? 
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Copyright © 19962018 Alexander BogomolnyLet point C lie on a segment AB. Draw (similarly oriented) semicircles with diameters AB, AC, and CB. The three semicircles enclose a shape, known as Arbelos  the shoemaker's knife  that has a long list of wonderful and unexpected properties. The applet attempts to suggest two of those.
Let D lie on the outer semicircle such that
 CEDF is a rectangle,
 EF is the common tangent to the two small semicircles.
What if applet does not run? 
Proof
Angles AEC, ADB, CFB are inscribed in respective semicircles and subtend their diameters. The three angles are therefore right. This implies property #1.
Let O_{a} be the center of the left semicircle. Triangle AO_{a}E is isosceles so that
∠AEO_{a} = ∠EAO_{a}. 
Angle ADC is complementary to both ∠DCE (in the right ΔCDE and ∠CAD (in the right ΔACD). Thus we also obtain
∠CAD ( = ∠EAO_{a}) = ∠DCE. 
Also, since CEDF is a rectangle,
∠CEF = ∠DCE, 
from where
∠CEF = ∠AEO_{a}. 
The last identity implies that angles AEC (which is right) and O_{a}EF are equal. The latter is therefore right.
Similarly, we can show that angle O_{b}FE is also right. Thus EF is indeed the common tangent to the two semicircles.
Remark
As was pointed out by Nathan Bowler, there is a simpler proof of the result. The part of the picture that consists of the circle AEC and the rectangle is symmetric with respect to the perpendicular bisector of the chord EC. In particular, the diagonal EF is the symmetric image of the diagonal CD. Since the latter is tangent to the circle at C, the former is tangent to the circle at E.
Another possibility is to consider the center O of the rectangle. Then
References

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Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny