Romanian Mathematical Olympiad - Final Round

ii) $\implies$ i): We have $|z-c| = |z-ta - (1-t)b| \le t|z-a| + (1-t)|z-b|$. Analogously, $|z-d| \le (1-t)|z-a| + t|z-b|$. Summing up, the conclusion follows.

i) $\implies$ ii): For $z = a$ we get $|a-b| \ge |a-c| + |a-d|$, and for $z = b$, $|a-b| \ge |b-c| + |b-d|$. Summing up, $2|a-b| \ge |a-c| + |a-d| + |b-c| + |b-d|$.
But $|a-c| + |b-c| \ge |a-b|$, $|a-d| + |b-d| \ge |a-b|$. It follows that all inequalities are in fact equalities. In particular $|a-c| + |b-c| = |a-b|$ giving the existence of $t_1 \in (0, 1)$ such that $c = t_1a + (1-t_1)b$. In the same way, there is $t_2 \in (0, 1)$ such that $d = t_2a + (1-t_2)b$.
We shall show that $t_1 + t_2 = 1$. We have $|a-c| + |b-c| = |a-b| = |b-c| + |b-d|$, implying $|a-c| = |b-d|$. These give $(1-t_1)|a-b| = t_2|b-a|$, which concludes the proof.