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Question

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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Solution

Let BD=CE=AF=x;BDF=CED=AFE=θ.
Note that AFD=B+θ,
and hence DFE=B. Similarly, EDF=CandFED=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
cxsinθ=kasinB=2Rkab,
where R is the circum-radius of ABC.
2Rksinθ=b(cx)/a.
Similarly, we obtain 2Rksinθ=c(ax)/band2Rksinθ=a(bx)/c.
We therefore get b(cx)a=c(ax)b=a(bx)c(1)
If some two sides are equal, say, a=b, then a(cx)=c(ax) 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(cx)<a(bx), form (1). Since b>aandcx>bx, we get b(cx)>a(bx), which is a contradiction.
Case 2. a<c<b––––––––––.
We may write (1) in the form
(cx)a/b=(ax)b/c=(bx)c/a(2).
Now a<c gives ax<cx 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.
284896_303204_ans.png

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