A Tangent in Concurrency
What is this about?
A Mathematical Droodle
What if applet does not run? 
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander BogomolnyThe 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 T_{A}, T_{X}, and T_{B} denote the tangents to the circle at A, X, and B respectively. Assume that BX and AX produced beyond X intersect T_{A} and T_{B} in Y and Z, respectively. Then the three lines T_{X}, YZ, and AB are either parallel or concurrent. 
What if applet does not run? 
Proof
Let T_{X} meets T_{A} in U and T_{B} 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 T_{X} also passes through P.
Note that UA and UX are tangent to the same circle and are therefore equal:
(1)  ∠UAX = ∠UXA. 
Further, the fact that angles YAB and AXY are right implies
(2) 

ΔXUY is therefore isosceles, and
(3)  UY = UX. 
Combining (3) with
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,
(I must note that the statement just proven is a particular case of a more general theorem.)
References
 E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995, #314
Activities Contact Front page Contents Geometry
Copyright © 19962018 Alexander Bogomolny