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

Explanation

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

1. XY bisects angle EYF.

2. GX = GY.

3. EO is orthogonal to FO.

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

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

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

There are two tangents to the circle from E: EX = AE. Similarly for the point B:

Assume first G is the intersection of AF and BE. Triangles AEG and BFG are similar. Therefore,

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

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)

Since GY = GX,

the harmonic mean of AE and BF.

Finally, since EOF = 90^o, the right triangles AEO and BFO are similar. From which

References

1. P. Y. Wu, Leon Bankoff's Surprises