| |||||||||||||||||||||||||||||||||||||
Let 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
ProofAngles AEC, ADB, CFB are inscribed in respective semicircles and subtend their diameters. The three angles are therefore right. This implies property #1. Let Oa be the center of the left semicircle. Triangle AOaE is isosceles so that
Angle ADC is complementary to both
Also, since CEDF is a rectangle,
from where
The last identity implies that angles AEC (which is right) and OaEF are equal. The latter is therefore right. Similarly, we can show that angle ObFE is also right. Thus EF is indeed the common tangent to the two semicircles. RemarkAs 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
|
| ||||||||||||||||||||||||||||||||||||