Using vectors how to prove $AB=AC$? 
In a triangle $ABC$, $D$ is the mid-point of the side $AB$ and $E$ is the centroid of triangle $CDA$. If $\vec{OE}\cdot\vec{CD}=0$, where $O$ is the circumcentre of triangle $ABC$, then using vectors, how to prove $AB=AC$?

I found out $\vec{OE}\cdot\vec{CD}=\frac14(3\vec a+\vec b+2\vec c)\cdot(\vec a+\vec b-2\vec c)=0$. After that how to do? I'm not able to reach the final answer.
$\vec a$, $\vec b$ and $\vec c$ are position vectors of $A$, $B$ and $C$, respectively.
 A: Let $O$ be the origin and let $\vec a$, $\vec b$, $\vec c$, $\vec d$ and $\vec e$, be equal to $\vec{OA}$, $\vec{OB}$, $\vec{OC}$, $\vec{OD}$ and $\vec{OE}$, respctively. By the definitions of $D$ and $E$, we have:
$$\vec d=\frac{\vec a+\vec b}2$$
$$\vec e=\frac{\vec a+\vec c+\vec d}3=\frac{3\vec a+\vec b+2\vec c}6$$
Now, since $\vec{CD}=\vec d-\vec c=\dfrac{\vec a+\vec b-2\vec c}2$, by $\vec{OE}\cdot\vec{CD}=0$ we get:
$$\left(3\vec a+\vec b+2\vec c\right)\cdot\left(\vec a+\vec b-2\vec c\right)=0$$
$$\therefore\quad3\vec a\cdot\vec a+\vec b\cdot\vec b-4\vec c\cdot\vec c+4\vec a\cdot\vec b-4\vec a\cdot\vec c=0\tag{*}\label{*}$$
But note that $OA$, $OB$ and $OC$ are equal to the circumradius, so we have $\vec a\cdot\vec a=\vec b\cdot\vec b=\vec c\cdot\vec c=R^2$ and thus $3\vec a\cdot\vec a+\vec b\cdot\vec b-4\vec c\cdot\vec c=0$. Hence by \eqref{*} we have $\vec a\cdot\left(\vec b-\vec c\right)=0$, i.e. ${OA}$ is perpendicular to $BC$. So the median $OA$ is the altitude too, which yields $AB=AC$.
