Concurrence

Add a point $C'$ symmetric to $C$ about $AB$ to get an equilateral triangle $\triangle{ACC'}$ . $BD$ is parallel to $AC'$ , as it connects the midpoints of $AC$ and $CC'$ , and because $\triangle{ACC'}$ is equilateral, $BD$ and $AC'$ share a common perpendicular bisector - $CD'$ . $C'D$ is a perpendicular bisector of $AC$ , so $AB$ , $C'D$ and $CD'$ all intersect at $E$ - the centroid/incenter/circumcenter/orthocenter of $\triangle{ACC'}$ .

Drop a perpendicular to $\overline{AC}$ from $D.$ Say it intersects $\overline{AB}$ at $E.$ Since $BCD$ is equilateral, the perpendicular from $C$ bisects $\overline{BD},$ so all that is required is to show that that perpendicular goes through $E,$ i.e. to show that $\overline{CE}$ is perpendicular to $\overline{BD}.$

To see this, consider right triangles $CDE$ and $CBE.$ These share a hypotenuse, $\overline{CE},$ and have congruent legs $\overline{CD}$ and $\overline{CB}.$ So they are congruent. Hence $\angle{DCE} = \angle{BCE} = \frac12 \angle{ACB} = 30^{\circ}.$ The fact that $\overline{CE}$ bisects $\angle{BCD}$ shows that $\overline{CE}$ is perpendicular to $\overline{BD}$ : e.g. if $\overline{CE}$ and $\overline{BD}$ intersect at $F,$ it's clear that $\angle{CFB} = 180^\circ - 30^\circ - 60^\circ = 90^\circ.$

Choose a coordinate system with A at (0,0). Then, by properties of 30-60-90 triangles, the coordinates of the other points are D(Sqrt(3)/2, 1/2), C(Sqrt(3), 1), B(Sqrt(3), 0). If L1 is the perpendicular bisector of AC, we find L1 has slope of -Sqrt(3). Since it passes through D, it's equation is L1(x) = 1/2 - Sqrt(3)*(x-Sqrt(3) / 2). This intersects AB when it's y-coordinate is 0. Doing the algebra yields x = 2/Sqrt(3). Similarly, we can find an equation for L2, the perpendicular bisector of DB. Setting that equation equal to zero also yields x = 2/Sqrt(3), which is the same coordinate as the zero of L1. Therefore, they do meet on AB (at a distance of 2/Sqrt(3) units from A.