Hint

Beautiful problem vlwk wrote: Let $E$ be a point outside $\triangle{ABC}$ such that $\triangle{ABD} \sim \triangle{ACE}$. We get $CE=BD\cdot \frac{AC}{AB}=cd,AE=AD\cdot \frac{AC}{AB}=\frac{acd}{b}$ We get $\angle{BAC}=\angle{DAE}$ and $\frac{AB}{AC}=\frac{b}{c}=\frac{AD}{AE}$ So $\triangle{ADE} \sim \triangle{ABC}$ give us $DE=cd$ So $\triangle{CDE}$ is equilateral triangle So $\angle{ABD}+\angle{ACD}=\angle{DCE}=60^{\circ}$

related: https://artofproblemsolving.com/community/c6h112625 this problem should have been posted in HSO forum instead cos it is an olympiad problem.

Here's my solution using mainly algebra instead of geometry. Using the cosine rule on point C, we obtain (ab)^2 = (ac)^2 + (bc)^2 - 2(ac)(bc)cos(C) Divide by c^2 on both sides to get (ab/c)^2=a^2+b^2-2ab(cos(C)) - which equals c^2 for a given triangle A'B'C' with this means that ab/c=c --> ab=c^2 this can be reapplied to points A and B to obtain the symmetric algebraic equations ab=c^2 bc=a^2 ac=b^2 We know that a^2+b^2+c^2 >= ab+ac+bc, equality occurring when a=b=c. We know that it is equal due to the trio of equations above, and a=b=c. This means that ABC with side lengths ab, bc, ac, are all the same and ABC is an equilateral triangle. From there we can just use perpendicular bisectors on AB, BC and AC to show that they intersect as well, to show that the intersection of points is Point D, which is both the incenter and circumcenter of triangle ABC. This means that the perpendicular bisector passing through its respective angle bisects it as well, and both angles ABD and ACD are 30 degrees. This means that angles (ABD+ACD)= 60 degrees. ...yes I don't know how to use LaTeX, how could you tell