Let $ABC$ be equilateral triangle and $D$ be a point on the arc BC of the circle circumscribing ABC. Then prove using complex algebra that : $|BD| +|CD| = |AD|$.
For me, the confusion is both : (i)- in ability to think of problem, and (ii)- to formulate it.
(i) Ability to think of problem : How is it possible that the two points nearest to $D$ have their arc lengths equal to that of $|AD|$, where A is the distant point. I cannot imagine by taking the shortest length of sum of $|BD| + |CD|$, or taking by a particular direction (cck, ck).
(ii) Unable to find a way to formulate, except possibly by taking the angle possible, and then would involve a range of $60^0$ for any point $D$ on the arc $BC$. Even then difficult to think of about the formulation.
Let $A(a)$, $B(b)$, $C(c)$ and $D(z)$.
Thus, by Ptolemy $$AB\cdot DC+BD\cdot AC=AD\cdot BC$$ or $$|a-b|\cdot |z-c|+|b-z|\cdot|a-c|=|a-z|\cdot |b-c|$$ and since $$|a-b|=|a-c|=|b-c|,$$ we obtain $$|z-c|+|b-z|=|a-z|$$ or $$CD+BD=AD.$$
Another way.
Let $A(1,0)$, $B(\cos120^{\circ}+i\sin120^{\circ})$, $C(\cos240^{\circ}+i\sin240^{\circ})$ and $D(\cos\phi+i\sin\phi)$,
where $120^{\circ}\leq\phi\leq240^{\circ}$.
Thus, we need to prove that $$\sqrt{(1-\cos\phi)^2+\sin^2\phi}=$$ $$=\sqrt{(\cos\phi-\cos120^{\circ})^2+(\sin\phi-\sin120^{\circ})^2}+\sqrt{(\cos\phi-\cos240^{\circ})^2+(\sin\phi-\sin240^{\circ})^2}$$ or $$\sqrt{2-2\cos\phi}=\sqrt{2-2\cos(120^{\circ}-\phi)}+\sqrt{2-2\cos(240^{\circ}-\phi)}$$ or $$\left|\sin\frac{\phi}{2}\right|=\left|\sin\left(60^{\circ}-\frac{\phi}{2}\right)\right|+\left|\sin\left(120^{\circ}-\frac{\phi}{2}\right)\right|$$ or
$$\sin\frac{\phi}{2}=\sin\left(\frac{\phi}{2}-60^{\circ}\right)+\sin\left(120^{\circ}-\frac{\phi}{2}\right),$$ which is an easy identity.