Let D,E,F be points on the sides BC,CA,AB, respectively, of a triangle ABC such that BD=CE=AF and ∠BDF=∠CED=∠AFE. Prove that ΔABC is equilateral.
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Let BD=CE=AF=x;∠BDF=∠CED=∠AFE=θ. Note that ∠AFD=B+θ, and hence ∠DFE=B. Similarly, ∠EDF=Cand∠FED=A. ∴EFD is similar to ABC We may take FD=ka,DE=kbandEF=kc, for some positive real constant k. Applying sine rule to triangle BDF, we obtain c−xsinθ=kasinB=2Rkab, where R is the circum-radius of ABC. ∴2Rksinθ=b(c−x)/a. Similarly, we obtain 2Rksinθ=c(a−x)/band2Rksinθ=a(b−x)/c. We therefore get b(c−x)a=c(a−x)b=a(b−x)c(1) If some two sides are equal, say, a=b, then a(c−x)=c(a−x) giving a=c; we get a=b=c and ABC is equilateral. Suppose no two sides of ABC are equal. We may assume a is the least. Since (1) is cyclic in a,b,c, we have to consider two cases : a<b<canda<c<b. Case 1. a<b<c––––––––––––. In this case a<c and hence b(c−x)<a(b−x), form (1). Since b>aandc−x>b−x, we get b(c−x)>a(b−x), which is a contradiction. Case 2. a<c<b––––––––––––. We may write (1) in the form (c−x)a/b=(a−x)b/c=(b−x)c/a(2). Now a<c gives a−x<c−x so that bc<ab. This gives b2<ac. But b>aandb>c, so that b2>ac, which again leads to a contradiction Thus Case 1 and Case 2 cannot occur. ∴a=b=c.