Twin Segments in Arbelos
Here is a problem (10895) from the 2003 American Mathematical Monthly. It was proposed by August Wendijk, Jersey City, NJ. (The solution below is by Thomas Hermann, Milford, OH):
Given a point C on the segment AB, erect semicircles on diameters AB, AC, and BC, all on the same side of AB. Let L be the line through C perpendicular to AB. Let S be the largest circle that fits into the region bounded by L and the semicircles on diameters AC and BC. Let D be the point of tangency between S and the semicircle on BC. Extend the diameter of S through D until it hits L at E. Prove that AC and DE have the same length. 
What if applet does not run? 
References
 August Wendijk; Thomas Hermann, American Mathematical Monthly, Vol. 110, No. 1. (Jan., 2003), pp. 6364.
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Copyright © 19962018 Alexander BogomolnyLet's introduce the following notations
 a the radius of the circle on AC as a diameter,
 G the center of the circle with diameter BC; b the radius of the circle,
 H the center of the circle with diameter AB; R the radius of the circle,
 F the center of the Archimedian twin in the problem (the largest circle that fits into the region bounded by L and the semicircles on diameters AC and BC); r  the radius of the the circle,
 M the projection of F on AB; y = FM,
 T the point of tangency of circles H(R) and F(r),
 w = DE.
We have to show that w = 2a.
What if applet does not run? 
Observe that by the construction,
 CM = r,
 HM = b  a  r,
 GF = b + r,
 FH = a + b  r.
Triangles FGM and FHM are rightangled. The Pythagorean theorem implies:
(b + r)² = y² + (b  r)² and (a + b  r)² = y² + (b  a  r)², 
from which
r = ab / (a + b). 
Triangles FGM and CEG are similar so that
(b + r) / (b  r) = (b + w) / b. 
From the latter
r = bw / (2b + w). 
Comparing the to expressions for r yields w = 2a, as needed.

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Copyright © 19962018 Alexander Bogomolny