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tototo
1st June 2008 - 10:37 PM

In the figure below, given an square ABCD of area S, E is the intersection of BD and the semicircle of diameter AD, and CF is tangent to the semicircle. If S1 is the area of triangle DEF


User posted image: User posted image
rpenner
2nd June 2008 - 03:19 AM
I think it works out simply if you define
G = midpoint of AD and then by definition
GFC is a right triangle
rpenner
2nd June 2008 - 05:24 AM
Then, you will want to use properties of triangles and semicircles to prove F is a distance (1/5)√S from AB and a distance (2/5)√S from AD.

Thus all the points are identified with exact analytic coordinates and everything becomes easy.
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