= Solution
In the integral defining the <Beta function>, put $t=(1+x)/2$ and then $u=x^2$. Symmetry gives
$$
\begin{aligned}
B(z,z)
&=2^{1-2z}\int_{-1}^{1}(1-x^2)^{z-1}\,dx\\
&=2^{2-2z}\int_0^1(1-x^2)^{z-1}\,dx\\
&=2^{1-2z}\int_0^1u^{-1/2}(1-u)^{z-1}\,du\\
&=2^{1-2z}B(z,\tfrac12).
\end{aligned}
$$
This proves the identity for $\operatorname{Re}z>0$. Both sides extend by <analytic continuation> to the same <meromorphic function>, equivalently by the <beta--gamma identity> and the <Gamma duplication formula>. \b[The identity holds meromorphically for every $z\in\mathbb C$; both sides are finite for $z\notin\{0,-1,-2,\ldots\}$ and have matching poles at the excluded points.]
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