Let $BC = AD = a$, $AB = CD = b$, $AC = d_1$, $BD = d_2$. Let for the definiteness $\angle ADC \ge 90^\circ$. Then $d_1 > a$, $d_1 > b$. Without loss of generality we suppose that $a \ge b$. Then $a+a \ge a+b > d_1 \ge d_2$. Exactly two cases are possible.

1. The diagonals have the different colors. If two adjacent sides $a$ and $b$ are painted the same color, then together with the diagonal of the same color they form the triangle congruent with (by three sides) either the triangle $ADB$ or the triangle $ADC$. Now let two opposite sides $a$ and $b$ have the same color with one of the diagonals. Then one can construct the triangle with these sides and the diagonal as its sides since $a+a > d_1$ and $a+a > d_2$.

2. Both the diagonals are painted the same color, say green. Then one of the side is also green, and the side equaled green side and two remaining sides are painted red. If the sides $a$, $a$, $b$ are painted the same color, then one can construct the triangle with $a$, $a$, $b$ as its sides. It follows from the inequalities $a+a \ge a+b > b$. Suppose that $d_1, d_2, a$ are painted green and $b, b, a$ are painted red. One cannot form a triangle from the first triple only if $d_1 \ge a+d_2$ or $d_2 \ge a+d_1$ (since the inequality $d_1+d_2 > a$ certainly holds); one cannot form a triangle from the second triple only if $a \ge 2b$. Let $d_1 \ge a+d_2$. Then

d_1 $\ge$ 2b + d_2, i.e. AC $\ge$ AB + BD + DC > AD + DC, a contradiction. Similarly, if $d_2 \ge a + d_1$, then $d_2 \ge 2b + d_1$, i. e. $BD \ge BA + CD + AC > BA + AD$, a contradiction.

Therefore, there always exist three segments of the same color such that one can construct the triangle with these segments as its sides.