Apollonius’ Identity inner product space $||z-x||^2+||z-y||^2=\frac{1}{2}||x-y||^2+2||z-\frac{x+y}{2}||^2$ 
I proved it by expanding both sides and i found both sides are equal. Are there any easy way to prove it?

In a Hilbert space you can use the Parallelogram law : 
$$\|a+b\|^2 + \|a-b\|^2 = 2(\|a\|^2 + \|b\|^2)$$
So : 
$$\begin{align}
&\|z - \frac{x+y}{2} \|^2\\
&= \|\frac{z}{2} - \frac{x}{2} + \frac{z}{2} - \frac{y}{2}\|^2\\
&= 2(\|\frac{z}{2} - \frac{x}{2}\|^2 + \|\frac{z}{2} - \frac{y}{2}\|^2) - \|\frac{y}{2} + \frac{x}{2}\|^2\\
&= \frac12(\|z - x\|^2 + \|z - y\|^2 - \frac{1}{4}\|y - x\|^2\\
&\square\end{align}$$
