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.
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Posted by Bractals
on 2018-02-20 19:07:58 |
Solution
