Let $x, p\neq q\in \mathbb{R}^n$.
We have that $d(p,x)=d(q,x)$, so $\|p-x\|=\|q-x\|$, so $(p-x)\cdot (p-x)=(q-x)\cdot (q-x)$.
I want to show that $$x-2\frac{(q-p)\cdot \left (x-\frac{1}{2}(p+q)\right )}{(q-p)\cdot (q-p)}(q-p)=x$$ We have that \begin{align*}x&-2\frac{(q-p)\cdot \left (x-\frac{1}{2}(p+q)\right )}{(q-p)\cdot (q-p)}(q-p) \\ & =x-2\frac{(q-p)\cdot \left (\frac{1}{2}x+\frac{1}{2}x-\frac{1}{2}(p+q)\right )}{(q-p)\cdot (q-p)}(q-p)\\ & =x-2\frac{(q-p)\cdot \left (\frac{1}{2}(x-p)+\frac{1}{2}(x-q)\right )}{(q-p)\cdot (q-p)}(q-p)\\ & =x-2\frac{\frac{1}{2}(q-p)\cdot (x-p)+\frac{1}{2}(q-p)\cdot(x-q)}{(q-p)\cdot (q-p)}(q-p) \\ & =x-\frac{(q-p)\cdot (x-p)+(q-p)\cdot(x-q)}{(q-p)\cdot (q-p)}(q-p)\end{align*} Is that correct so far? How can we use the given property to get the result?