Two concentric circles of radius R and r, with R>r are both intersected by the same secant line. The points of intersection, in order, are A,B,C,D.
Prove AC*CD is constant.
Let O be the center of the two circles.
Let Q be the foot of the perpendicular
from O to the secant ABCD.
Let x = OQ
z = QC
Z = QA = QD
Then z^2 = r^2  x^2
Z^2 = R^2  x^2
AC = AQ + QC = Z + z
CD = QD  QC = Z  z
AC*CD = Z^2  z^2
= (R^2  x^2)(r*2  x^2)
= R^2  r^2
a constant
QED
Note: If the secant passes through the
center O, then the problem is trivial.

Posted by Bractals
on 20180220 19:07:58 