 Solution

Use relative homology. Every simplex of $X=B\cup C$ lies in $B$ or $C$, and those lying in both are exactly the simplices of $A$. Consequently inclusion induces a chain-complex isomorphismIt is bijective on the respective bases of simplices and commutes with boundary because it comes from inclusion. Thus

For any pair $(Y,Z)$, the long exact sequence in homology shows that inclusion induces isomorphisms in every degree exactly when $H_j(Y,Z)=0$ for every $j$. To verify both directions, if all relative groups vanish, exactness makes each $H_j(Z)\to H_j(Y)$ injective and surjective. Conversely, if the inclusion maps are isomorphisms, the map from $H_j(Y)$ into the relative group is zero, while the relative connecting map is injective and has zero image because the next inclusion is injective; hence the relative group itself is zero. This includes degree zero, with the negative-degree ordinary group zero.

Apply the criterion first to $(B,A)$ and then to $(X,C)$ using the chain isomorphism. It proves