SeeSaw Lemma: What is this about?
A Mathematical Droodle
What if applet does not run? 
Activities Contact Front page Contents Geometry
Copyright © 19962005 Alexander BogomolnyThe applet presents a diagram with several interesting properties. Let X be a point on a semicircle with diameter AB and center O. The tangent to the semicircle at X meets the tangents at A, B in E and F, as in the diagram. Let XY be perpendicular to AB. Then
AF, BE, XY are concurrent (in, say, point G.)
XY bisects angle EYF.
GX = GY.
EO is orthogonal to FO.
XY is the harmonic mean of AE and BF.
OA (=OB) is the geometric mean of AE and BF.
(I learned of this configuration and its properties from P. Y. Wu online resource.)
What if applet does not run? 
There are two tangents to the circle from E: EX = AE. Similarly for the point B:
(1) 
EX = AE, FX = BF. 
Assume first G is the intersection of AF and BE. Triangles AEG and BFG are similar. Therefore,
AG/GF = AE/BF = EX/FX. 
It follows that triangles AEF and GXF are also similar. Hence, GXAE. By construction,
(Virtually the same configuration presents a different case of concurrency. As it happened there, the present case corresponds to a more general situation by moving in the latter one of the points to infinity.)
Further, AY/BY = AE/BF. Therefore, triangles AYE and BYF are similar, so that
∠AYE = ∠BYF. 
Angles EYX and FYX being complementary to the above are also equal.
Next, we have two pairs of similar triangles: AGY is similar AFB and EGX is similar to EBF, wherefrom
(2) 
GY/BF = AY/AB and also GX/BF = EX/EF = AY/AB. 
Thus, GX = GY.
Concerning the orthogonality of EO and FO, note that triangles OXF and OBF are equal, so that FO is the bisector of angle BOX. Similarly, EO is the bisector of angle AOE. Which shows that
From (2)
(2') 

Since GY = GX,
XY = 2·AE·BF / (AE + BF), 
the harmonic mean of AE and BF.
Finally, since ∠EOF = 90°, the right triangles AEO and BFO are similar. From which
AE/AO = BO/BF. 
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Copyright © 19962018 Alexander Bogomolny64477696 