Proving that a map is a weak homotopy equivalence

Consider the category of pointed topological spaces $C$. Suppose objects $a,b,c,d,e,f,g,h \in C$. Suppose we also have the commuting diagrams where all the morphisms are serre fibrations $$\begin{array}& &a&\to &b\\ &\downarrow & &\downarrow \\ &c &\to &d\end{array}$$ $$\begin{array}& &e&\to &f\\ &\downarrow & &\downarrow \\ &g &\to &h\end{array}$$ and maps $f_{ae}:a \to e,f_{bf}:b \to f, f_{cg}:c \to g,f_{dh}:d \to h$ satisfying the obvious commutativity conditions. Then I want to show that if any three of the maps $f_{ae},f_{bf}, f_{cg},f_{dh}$ are weak homotopy equivalence then so is the other one. If anyone has any hints or suggestions it would be great. Thanks.

• @NajibIdrissi I meant any three of $f_{ae},f_{bf}, f_{cg},f_{dh}$ are weak homotopy equivalences – happymath Jun 17 '16 at 8:43

Is this true? If we take $a = b = \ldots = S^1$, and $f: S^1 \to S^1; z \mapsto z^2$, then we have
$$\require{AMScd} \begin{CD} S^1 @>id>> S^1 \\ @VidVV @VVidV \\ S^1 @>>id> S^1 \end{CD}$$
$$\require{AMScd} \begin{CD} S^1 @>id>> S^1 \\ @VidVV @VVfV \\ S^1 @>>f> S^1 \end{CD}$$
both commuting, and all maps being Serre fibrations ($f: S^1 \to S^1$ is a covering map of degree 2).
But we can then take $f_{ae}, f_{bf}, f_{cg} = id$, and $f_{dh} = f$ (i.e. all the maps between the diagrams are the identity except in the bottom-right corner) then we get a commutative cube, with three of the maps between diagrams being the identity, and the fourth not a weak homotopy equivalence.