Question -
equilateral triangles DBC, ECA and FAB are constructed externally on the sides BC,CA,AB respectively of triangle ABC. Prove that AD, BE, and CF are also concurrent.
My attempt -
I know this is fermat point and also that this question is already been answered but I am not getting how to proof this using ceva theorem to prove concurrency....I know the proof using rotation... kindly help me thanks ...
Let $A'$ be the intersection of $AD$ and $BC$. Then (if all angles are less than $2\pi/3$):
$$\frac{A'B}{A'C} = - \frac{\text{Area}(ABD)}{\text{Area}(ACD)} = - \frac{AB \cdot BD \cdot \sin{(B + \pi/3})}{AC\cdot CD \cdot \sin{(C + \pi/3)}} = - \frac{AB \sin{(B + \pi/3)}}{AC\sin{(C + \pi/3)}}.$$
Write the other two similar ratios and multiply to get -1. The equality between the first and the last term remains valid for angles exceeding $2\pi/3$.