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From Proposition
21 we have the following :
(BB'+CC'+...)/AA'
= A'B/ BA then we have
AB(BB' +CC'
+ .....) = AA'. A'B
From Proposition
29 we have the following :
(Radius of R) = Sqrt(AA'.A'B) > A'B
we know A'B = 20
P where P is the point in which AB touches the circle ama'm'.
so (radius R) > (diameter of circle ama'm')
finally we get
that the surface of the circumscribed solid is greater than
4 times the great circle of the given sphere .
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