Three circles are located in the plane such that each pair of circles
intersect in two points, thereby defining a common secant.

Prove that if two of the secants intersect, then all three secants are
concurrent.
  

Let the circles be C1, C2, C3, with PQ being the common
chord of C1 & C2, and RS the common chord of C1 & C3.
Given that the secants PQ and RS intersect at X, let T be
one of the intersections of C2 & C3 and let TX intersect
C2 and C3 elsewhere at points U & V respectively.

Using the intersecting chord theorem

in C1:   |PX||QX| = |RX||SX|                  (1)

in C2:   |PX||QX| = |TX||UX|                   (2)

in C3:   |RX||SX| = |TX||VX|                  (3)

Using (1), (2) and (3), |UX| = |VX|          (4)

Since X must lie either inside both C1 & C2 or outside both
C1 & C2, it follows that U and V must both be on the same
side of X, so U and V are coincident, from (4).
Thus TU and TV are the same common chord of C2 & C3,
proving that the three secants are concurrent (at X).

Posted by Harry on 2015-08-18 11:17:56