Prove or disprove the implication: Prove or disprove the implication:
$a^2\cdot \tan(B-C)+ b^2\cdot \tan(C-A)+ c^2\cdot \tan(A-B)=0 \implies$
$ ABC$ is an isosceles triangle.
I tried to break down the left hand side in factors, but all efforts were in vain.
Does anyone have a suggestion? 
Thank you very much!

RTP:$$                                                                                                        
a^{2}\tan(B-C)+b^{2}\tan(C-A)+c^{2}\tan(A-B)=0$$ $\implies$ ABC is an isosceles triangle                
$$                                                                                                        
c^{2}=b^{2}({\frac{\sin C}{\sin B}})^{2} \implies b^{2}=c^{2}({\frac{\sin B}{\sin C}})^{2}                
$$
$$                                                                              a^{2}=c^{2}({\frac{\sin(\pi-A)}{\sin C}})^{2}=c^{2}({\frac{\sin A}{\sin C}})^{2}                          
$$
$$
a^{2}\tan(B-C)+b^{2}\tan(C-A)+c^{2}\tan(A-B)=0
$$
$$
\implies c^{2}[({\frac{\sin A}{\sin C}})^{2}({\frac{\tan B-\tan C}{1+\tan B\tan C}})+({\frac{\sin B}{\sin C}})^{2\
}({\frac{\tan C-\tan A}{1+\tan C\tan A}})+({\frac{\tan A-\tan B}{1+\tan A\tan B}})]=0
$$
$$
\implies [({\frac{\sin A}{\sin C}})^{2}({\frac{\tan B-\tan C}{1+\tan B\tan C}})+({\frac{\sin B}{\sin C}})^{2\
}({\frac{\tan C-\tan A}{1+\tan C\tan A}})+({\frac{\tan A-\tan B}{1+\tan A\tan B}})]=0
$$
Now for any A>B>C the above equation will be false. [It is a conjecture]
Conjecture is a statement which is accepted as to be true/false until any counter example is presented.
So now the above equation is false (for A>B>C) according to conjecture.
Now there are only two possibilities either any two angles be equal or all three angles be equal.
In both the cases the above equation will be true.
and therefore the implication is proved to be right.
