See-Saw Lemma: What is this about?
A Mathematical Droodle

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Explanation

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Copyright © 1996-2005 Alexander Bogomolny

The 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

  1. AF, BE, XY are concurrent (in, say, point G.)

  2. XY bisects angle EYF.

  3. GX = GY.

  4. EO is orthogonal to FO.

  5. XY is the harmonic mean of AE and BF.

  6. OA (=OB) is the geometric mean of AE and BF.

(I learned of this configuration and its properties from P. Y. Wu online resource.)

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at https://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.

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, GX||AE. By construction, AE ⊥ AB. But, since also XY ⊥ AB, we see that indeed G lies on XY.

(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 EO ⊥ FO.

From (2)

(2')
GY= BF·AY/AB
 = BF·EX/EF
 = BF·AE/EF.

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 © 1996-2018 Alexander Bogomolny

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