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There is another way to solve, let F be the diagonal opposite point B of the triangle ABF right at A, from which we have BC*BF=AB², from which 2BO*BC=64, so BD*BC=32, which is the area of the rectangle.
There is another way to solve, let F be the diagonal opposite point B of the triangle ABF right at A, from which we have BC*BF=AB², from which 2BO*BC=64, so BD*BC=32, which is the area of the rectangle.