Question

Let $D,E,F$ be points on the sides $BC,CA,AB,$ respectively, of a triangle $ABC$ such that $BD=CE=AF$ and $\angle BDF=\angle CED=\angle AFE$. Prove that $\Delta ABC$ is equilateral.

Answer 1

Let $BD=CE=AF=x;\angle BDF=\angle CED=\angle AFE=\theta$.
Note that $\angle AFD=B+\theta$,
and hence $\angle DFE=B$. Similarly, $\angle EDF=Cand\angle FED=A$. $\therefore EFD$ 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
$\frac{c-x}{sin\theta }=\frac{ka}{sinB}=\frac{2Rka}{b}$,
where $R$ is the circum-radius of $ABC$.
$\therefore 2Rksin\theta =b(c-x)/a$.
Similarly, we obtain $2Rksin\theta =c(a-x)/band2Rksin\theta =a(b-x)/c$.
We therefore get $\frac{b(c-x)}{a}=\frac{c(a-x)}{b}=\frac{a(b-x)}{c}\left(1\right)$
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 $\left(1\right)$ is cyclic in $a,b,c$, we have to consider two cases : $a<b<canda<c<b$.
Case 1. $\underset{–}{a<b<c}$.
In this case $a<c$ and hence $b(c-x)<a(b-x)$, form $\left(1\right)$. Since $b>aandc-x>b-x$, we get $b(c-x)>a(b-x)$, which is a contradiction.
Case 2. $\underset{–}{a<c<b}$.
We may write $\left(1\right)$ in the form
$\frac{(c-x)}{a/b}=\frac{(a-x)}{b/c}=\frac{(b-x)}{c/a}\left(2\right)$.
Now $a<c$ gives $a-x<c-x$ so that $\frac{b}{c}<\frac{a}{b}$. This gives $b^2<ac$. But $b>aandb>c$, so that $b^2>ac$, which again leads to a contradiction
Thus Case $1$ and Case $2$ cannot occur.
$\therefore a=b=c$.