About a set of four concurrent circles (circles passing through a one common point) of same radius r,four of the common tangents are drawn to determine the circumscribing quadrilateral ABCD.Prove that ABCD itself is a cyclic quadrilateral.
It's certainly true if all the tangents are drawn exterior to the group of circles.
Assuming this is what is meant by 'circumscribing quadrilateral':
Call the centres of the circles O1,O2,O3,O4.
Then quadrilateral O1O2O3O4 is cyclic, because the one common point, p, is exactly one radius away from each of O1,O2,O3,and O4.
Now line O1O2, for example, connects two of the centres of the circles. The common tangent is parallel to this line, hence the points of tangency are each exactly one more perpendicular radius away from O1 and O2.
The result, after the same procedure is completed for the remaining 3 sides of O1O2O3O4, is that the angles of the external quadrilateral ABCD are the same as those of O1O2O3O4.
Accordingly, ABCD is itself also a cyclic quadrilateral.
 Over to you, Bractals, for the rest of it.
Edited on August 23, 2014, 11:34 pm

Posted by broll
on 20140823 11:45:34 