I don't know the name of the following theorem, or even if it has a name.
If $P(A) = P(B)$, and $P(A | C) = P(B | C)$, then $P(C | A) = P(C | B)$
Proof. Suppose (i) $P(A) = P(B)$ and (ii) $P(A | C) = P(B | C)$. Then: \begin{align} \frac{P(A\wedge C)}{P(C)} &= \frac{P(B\wedge C)}{P(C)}\quad \textrm{(From (ii))}\\ P(A\wedge C) &= P(B\wedge C)\\ \frac{P(A\wedge C)}{P(A)} &= \frac{P(B\wedge C)}{P(A)}\\ \frac{P(A\wedge C)}{P(A)} &= \frac{P(B\wedge C)}{P(B)}\quad \textrm{(From (i))}\\ P(C|A) &= P(C|B) \end{align} Can anyone enlighten me?
It's essentially an implication of Bayes' theorem. Assuming the various probabilities are nonzero, which you do implicitly,
$$\frac{\Pr(C|B)}{\Pr(C|A)} = \frac{ \frac{\Pr(B|C) \Pr(C)}{\Pr(B)}}{\frac{\Pr(A|C) \Pr(C)}{\Pr(A)}}=1. $$