A Tangent in Concurrency
What is this about?
A Mathematical Droodle


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Explanation

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Copyright © 1996-2018 Alexander Bogomolny

The applet purports to illustrate the following statement:

  Let AXB be a semicircle with diameter AB and center O (X being a point on the semicircle different from A and B.) Let TA, TX, and TB denote the tangents to the circle at A, X, and B respectively. Assume that BX and AX produced beyond X intersect TA and TB in Y and Z, respectively. Then the three lines TX, YZ, and AB are either parallel or concurrent.


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What if applet does not run?

Proof

Let TX meets TA in U and TB in V.

The three lines are clearly parallel if X lies in the middle of the arc AB. Assuming it is not, AU and BV are not equal, so that UV does meet AB. Let P be the point of intersection. I'll show that TX also passes through P.

Note that UA and UX are tangent to the same circle and are therefore equal: UA = UX. ΔAUX is isosceles, so that

(1) ∠UAX = ∠UXA.

Further, the fact that angles YAB and AXY are right implies

(2)
∠UYX = ∠AYX
  = 90O - ∠YAX
  = 90O - ∠UXA
  = ∠UXY

ΔXUY is therefore isosceles, and

(3) UY = UX.

Combining (3) with UA = UX we see that U is the midpoint of AY. Similarly, V is the midpoint of BV. On the other hand, if UP meets BZ in V', then, considering several pairs of similar triangles (the ones cut on the lines UP, AB, and YZ by the two parallels TA and TB), we also get V'B = V'Z. Hence V' = V, which proves the statement.

We might have started a little differently. Assume, for example, that U' is the midpoint of AY. Then XU' is a median to the hypotenuse of the right triangle AXY, which, as well known, means that

(4) U'A = U'Y = U'X.

Since U'A = U'X, and U'A is tangent to the circle, so is U'X. But the is only one tangent to the circle at X. Therefore, U = U'.

(I must note that the statement just proven is a particular case of a more general theorem.)

References

  1. E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995, #314
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Copyright © 1996-2018 Alexander Bogomolny
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