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This question originates from another one of a similar name: area of square between tangent(externally) circles

I want to know that if two circles having the same radius are externally tangent, the tangent line between them divides the square (whose top two vertices are touching one circle each) lying on the same base as the circles into two equal rectangles. Moreover, we have to prove that the side length of the square is less than the radius of any of the two circles.


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A diagram can help.Can you post a diagram too? – iostream007 Jun 27 '13 at 18:12
Tried to upload diagrams as and but both appear broken. – MvG Jun 28 '13 at 12:38
@iostream,you can find it from MvG's comment. – rahul Jun 28 '13 at 15:23
A better wording:There are two circles,each of equal radius,that are externally tangent to one another.There is a small square between those circles,whose top two vertices are touching the circumference of one circle each.The base of the square lies on a common tangent.We have to prove that the tangent line between them divides the square into two equal rectangular regions. – rah4927 Dec 18 '13 at 14:19
I think that both statements are obvious: The first because of the symmetry of the situation and the second from a glance at the figure. – Christian Blatter Dec 18 '13 at 14:46

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