Indija mo 2011

Consider the triangle $BDF$, $CED$ and $AFE$ with $BD$, $CE$ and $AF$ as bases. The sides $DF$, $ED$ and $FE$ make equal angles $\theta$ with the bases of respective triangles. If $B\ge C\ge A$, then it is easy to see that $FD\ge DE\ge EF$. Now using the triangle $FDE$, we see that $B\ge C\ge A$ gives $DE\ge EF\ge FD$. Combining, you get $FD=DE=EF$ and hence $A=B=C=60^{\circ }$.