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


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?

Explanation

|Activities| |Contact| |Front page| |Contents| |Geometry|

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.


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?

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

Symmedian

  1. All about Symmedians
  2. Symmedian and Antiparallel
  3. Symmedian and 2 Antiparallels
  4. Symmedian in a Right Triangle
  5. Nobbs' Points and Gergonne Line
  6. Three Tangents Theorem
  7. A Tangent in Concurrency
  8. Symmedian and the Tangents
  9. Ceva's Theorem
  10. Bride's Chair
  11. Star of David
  12. Concyclic Circumcenters: A Dynamic View
  13. Concyclic Circumcenters: A Sequel
  14. Steiner's Ratio Theorem
  15. Symmedian via Squares and a Circle
  16. Symmedian via Parallel Transversal and Two Circles
  17. Symmedian and the Simson
  18. Characterization of the Symmedian Point with Medians and Orthic Triangle
  19. A Special Triangle with a Line Through the Lemoine Point

|Activities| |Contact| |Front page| |Contents| |Geometry|

Copyright © 1996-2018 Alexander Bogomolny

 63707777

Search by google: