If $ A\subset X$ is retract of $X$ and the inclusion $i :A\to X$ is a cofibration, prove that: $$\widetilde{H}_n(X)\equiv \widetilde{H}_n(A)\bigoplus\widetilde{H}_n(X/A).$$
where $X/A$ is quotient space, and $\widetilde{H}_n$ is the reduced homology group.
I hope to use the theorem: if $i :A\to X$ is a cofibration then there is a long exact sequence of reduced homology group and math induction to prove it, but I do not know how to use retract condition.
Recall that a short exact sequence $$0\xrightarrow{\ \ \ }A\xrightarrow{\ f \ }B\xrightarrow{\ g \ }C\xrightarrow{\ \ \ }0$$ splits if and only if there exists a map $p:B\to A$ such that $p\circ f=id_A$. Now, as you said, since the inclusion $i:A\hookrightarrow{X}$ is a cofibration, the projection $p:(X,A)\to(X/A,\ast)$ induces an isomorphism $H_n(X,A)\cong \tilde{H}_n(X/A)$, and therefore the long exact sequence of the pair $(X,A)$ becomes: $$\cdots\xrightarrow{\ \ \ }\tilde{H}_{n+1}(X/A)\xrightarrow{\ \partial \ }\tilde{H}_n(A)\xrightarrow{\ i_* \ }\tilde{H}_n(X)\xrightarrow{\ p_* \ }\tilde{H}_n(X/A)\xrightarrow{\ \partial \ }\tilde{H}_{n-1}(A)\xrightarrow{\ \ \ }\cdots$$ The condition $r\circ i=id_A$ (since $r$ is a retraction) implies that $r_*\circ i_*=id_*$ by functorality, and therefore $i_*$ is injective. You can show that this implies that the above reduces to the short exact sequence $$0\xrightarrow{\ \ \ }H_n(A)\xrightarrow{\ i_* \ }H_n(X)\xrightarrow{\ j_* \ }H_n(X,A)\xrightarrow{\ \ \ }0$$ By the above fact about splitting, this splits, giving $\tilde{H}_n(X)\cong \tilde{H}_n(A)\oplus \tilde{H}_n(X/A)$.