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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 746 details.
Triangle ABE similar to triangle FCB …. ( case AA)So b/a= BF/ABTriangle ABD similar to triangle AFC….( Case AA)So b/x=AF/AB=(AB+BF)/AB=1+BF/AB=1+b/aAnd x=ab/(a+b)
x/a + x/b = DC/AC + AD/AC = 1x = ab/(a + b)