This sounds strange to me: either some condition is missing, or the result should be $\angle ABC + \angle ACD \leq \frac{\pi}{3}$. Anyway, D is the 1st isodynamic point of the triangle ABC. Darij

Oops, sorry, I typed it wrongly. It should be $\angle ABD + \angle ACD = \frac{\pi}{3}$.

hmm, it seems to me that you either edited your post, or you just re-posted the same thing again

I edited the main post and added an extra note. If you see what Darij wrote you'll notice where I made the typo.

Aha, now clear. Not intending to take the game away from the beginners, I just would like to note that I've already said a lot noticing that D is the 1st isodynamic point of triangle ABC. Darij

Then it's obvious that the isogonal conjugate of D is the Fermat point of ABC

There is another solution, not using isogonal conjugates: It is easy to show (from the AD : BD : CD ratios) that the pedal triangle of D is equilateral. Then, a simple angle chase reveals the required result. Only the complete blockhead has not noted yet the empiric fact that one half of all IMO / IMO Shorlist problems on triangle geometry have to be solved using pedal triangles. I actually can't help wondering whether there are some big fans of pedal triangles in the jury or the guys there have no better ideas Darij

Invert around D (X' is inverted point of X) triangle A'B'C' is equilateral. Problem is very obvious.

Oh darij grinberg my solution is the same as yours ,excuseme I didn't read your solution