Here, $A C=B C=2 a$

$D$ and $E$ are the midpoints of $B C$ and $AC$.

$\therefore A E=E C=a$ and $B D=D C=a$

$\operatorname{In} \Delta A D C,(A D)^{2}=(A C)^{2}-(D C)^{2}$

$=(2 a)^{2}-(a)^{2}=4 a^{2}-a^{2}=3 a^{2}$

$A D=a \sqrt{3}$

Similarly, potential at point $D$ due to the given charge configuration is

$V_{D}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{B D}+\frac{q}{D C}+\frac{q}{A D}\right]$

$=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{a}+\frac{1}{a}+\frac{1}{\sqrt{3} a}\right]=\frac{q}{4 \pi \varepsilon_{0} a}\left[2+\frac{1}{\sqrt{3}}\right]………(i)$

Potential at point $E$ due to the given charge configuration is $V_{E}=\frac{1}{4 \pi \varepsilon_{0}}\left[\frac{q}{A E}+\frac{q}{E C}+\frac{q}{B E}\right]$

$=\frac{q}{4 \pi \varepsilon_{0}}\left[\frac{1}{a}+\frac{1}{a}+\frac{1}{a \sqrt{3}}\right]=\frac{q}{4 \pi \varepsilon_{0} a}\left[2+\frac{1}{\sqrt{3}}\right]………(ii)$

From the $(i)$ and $(ii)$, it is clear that

$V_{D}=V_{E}$

The work done in taking a charge $Q$ from $D$ to $E$ is

$W = Q({V_E} – {V_D}) = O$ $(\because \,{V_D} = {V_E})$