Maximum distance from closest vertex of rhombus

Question

Consider the unit rhombus formed by joining following coordinates $A(0,0), B(1,0), C(\frac{3}{2}, \frac{\sqrt{3}}{2}), D(\frac{1}{2}, \frac{\sqrt{3}}{2})$

What is the largest possible distance from nearest vertex of a given point $P$ inside rhombus (and what will be that point)?

It appears that the answer should be $\frac{1}{2}$ and point should be the mid point of rhombus. A detailed analysis would be appreciated.

Motivation for this question was the fact that maximum distance of any point inside a unit square from its closest vertex is $\frac{1}{\sqrt{2}}$ (which is for the mid point of square).

Answer 1

This rhombus is formed by two equilateral triangles. There are two points of maximum distance from the vertices, which are the circumcenters of triangles $ABD$ and $BCD$. The distance from either point to the nearest vertex is $\sqrt3/3$, which is greater than $1/2$.

Answer 2

In a slightly more general setting, the partition of a rhombus $ABCD$ into Voronoi cells just depends on the location of $X$ and $Y$, where $X$ and $Y$ are the intersections of the perpendicular bisectors of adjacent sides:

and the critical distance is $$ XA=XB=XC=YC=YB=YD=\color{red}{\frac{BC}{\sqrt{3}}}.$$