Specific identity involving gamma function

Question

I was trying to obtain Bailey's hypergeometric summation formula relying on this paper. The formula is:
$$_{2}F_{1}(a,1-a;c;\frac{1}{2}) = \frac{\Gamma(\frac{c}{2})\Gamma(\frac{c+1}{2})}{\Gamma(\frac{c+a}{2})\Gamma(\frac{c-a+1}{2})}.$$
However, using Euler's hypergeometric transformation and Beta function, I've got the following:
$$_{2}F_{1}(a,1-a;c;\frac{1}{2}) = \frac{2^{a-1} \Gamma(c) \Gamma(\frac{c+a-1}{2})}{\Gamma(c+a-1) \Gamma(\frac{c-a+1}{2})}.$$
I've checked those two expression for different values of $a$ and $c$ in Wolfram Alpha, and they seem to agree. I'm not very skillful with special functions, so I don't know how to prove that they're equal (or wrong?).
I'd appreciate any help.
Edit: perhaps it requires a proper application of Legendre's duplication formula

Answer 1

Yes. It follows from Legendre's duplication formula: $$\Gamma(2z)=2^{2z-1}\frac{\Gamma(z)\Gamma\left(z+\frac12\right)}{\sqrt \pi}.$$ A direct application of the formula gives that \begin{align*} \frac{2^{a-1} \Gamma(c) \Gamma(\frac{c+a-1}{2})}{\Gamma(c+a-1) \Gamma(\frac{c-a+1}{2})}&=2^{a-1}\frac{\Gamma(\frac{c+a-1}{2})}{\Gamma(\frac{c-a+1}{2})}\cdot\frac{2^{c-1}\Gamma\left(\frac{c}2\right)\Gamma\left(\frac{c+1}2\right)\sqrt\pi}{2^{c+a-2}\Gamma\left(\frac{c+a-1}2\right)\Gamma\left(\frac{c+a-1+1}2\right)\sqrt\pi}\\ &=\frac{\Gamma(\frac{c}{2})\Gamma(\frac{c+1}{2})}{\Gamma(\frac{c+a}{2})\Gamma(\frac{c-a+1}{2})}. \end{align*}