Let $A_2$ and $C_2$ be the reflections of $A_1$ and $C_1$ across $AB$ and $BC$ respectively. We see that $\angle BA_2A = \angle BA_1A = 180^{\circ} - \angle A_1CA = 180^{\circ} - \angle BCA$ and $\angle BC_2C = \angle BC_1C = 180^{\circ} - \angle C_1AC = 180^{\circ} - \angle BAC$, so $A_2$ and $C_2$ lie on $(ABCD)$. Also, $\angle DA_2B = \angle DAB = 3\angle ABC = \angle C_1BA_1 + \angle ABC + \angle A_1BC_2 = \angle A_2BA + \angle ABC + \angle CBC_2 = \angle A_2BC_2$, so $DA_2BC_2$ is an isosceles trapezoid with $BD = A_2C_2$. Thus, by the triangle inequality, $3A_1C_1 = A_2C_1 + A_1C_1 + A_1C_2 > A_2C_1 + C_1C_2 > A_2C_2 = BD$.

Woah....magical solution