= 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 <isomorphism>
$$
C_*(B)/C_*(A)\ \cong\ C_*(X)/C_*(C).
$$
It is bijective on the respective bases of simplices and commutes with boundary because it comes from inclusion. Thus
$$
H_j(B,A)\cong H_j(X,C)\quad\text{for every }j.
$$
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
$$
\boxed{H_*A\longrightarrow H_*B\text{ is an isomorphism}
\ \Longleftrightarrow\
H_*C\longrightarrow H_*X\text{ is an isomorphism}}.
$$
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