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Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
Let S be the area. Let BC=a, AC=b, AB=c. Since ΔABC~ΔDBE, thus DE/AC=BD/AB. d / b = (c sinA sinC) / cd = b sinA sinC = b (a/2R) (c/2R)Hence, S = abc/4R = Rd.
D,F,E,B are concyclic and they lie on the circle on BF as diameter.In this circle d = chord EF = (diameter BF) x sin B (i.e.) d = h Sin B where h denotes height BF.In ΔABC we have 2Δ= h.b = h. 2R sin B = 2R .h Sin B = 2R.dHence Δ = R.d
R=BC/2*BA/BFd=AC*BD/BARd=BC*AC*BD/(2BF)Due to similar triangles BC*BD=BF^2Therefore, Rd=AC*BF/2=[ABC]