Question

Let $ABC$ be an acute-angled triangle triangle, and let $D,E,F$ be points on $BC,CA,AB$ respectively such that $AD$ is the median, $BE$ is the internal angle bisector and $CF$ is the altitude. Suppose $\angle FDE=\angle C,\angle DEF=\angle Aand\angle EFD=\angle B$. Prove that $ABC$ is equilateral.

Answer 1

Since $\Delta BFC$ is right-angled at $F$, we have $FD=BD=CD=a/2$. Hence $\angle BFD=\angle B$. Since $\angle EFD=\angle B$, we have $\angle AFE=\pi -2\angle B$. Since $\angle DEF=\angle A$, we also get $\angle CED=\pi -2\angle B$. Applying sine rule in $\Delta DEF$, we have
$\frac{DF}{sinA}=\frac{FE}{sinC}=\frac{DE}{sinB}$.
Thus we get $FE=c/2andDE=b/2$. Sine rule in $\Delta CED$ gives
$\frac{DE}{sinC}=\frac{CD}{sin(\pi -2B)}$.
Thus $\left(b/sinC\right)=\left(a/2sinBcosB\right)$. Solving for $cosB$, we have
$cosB=\frac{asinc}{2bsinB}=\frac{ac}{2b^2}$.
Similarly, sine rule in $\Delta AEF$ gives
$\frac{EF}{sinA}=\frac{AE}{sin(\pi -2B)}$.
This gives (since $AE=b/c(a+c)$), as earlier,
$cosB=\frac{a}{a+c}$.
Compairing the two values of $cosB$, we get $2b^2=c(a+c)$. We also have
$c^2+a^2-b^2=2cacosB=\frac{2a^{2}c}{a+c}$.
Thus
$4a^{2}c=(a+c)(2c^2+2a^2-2b^2)=(a+c)(2c^2+2a^2-c(a+c\left)\right)$.
This reduces to $2a^3-3a^{2}c+c^3=0$. Thus $(a-c{)}^2(2a+c)=0$. We conclude that $a=c$. Finally
$2b^2=c(a+c)=2c^2$
We thus get $b=c$ and hence $a=c=b$. This shows that $\Delta ABC$ is equilateral.