Let $h \colon A \to B$ and $r \colon S^{n-1} \to A$ be continuous maps. Assume that $h$ is an homotopy equivalence, prove that $$ D^n \cup_{r} A \simeq D^n \cup_{h \circ r} B$$
where $D^n \cup_{r} A$ is the push out of the square
And the second is analogue.
My attempt
Being $h$ an homotopy equivalence implies the existence of a map $s \colon B \to A$ such that $A\circ B \simeq Id$ $B\circ A \simeq Id$.
We already proved that if $a \simeq b$ maps from $S^{n-1} \to X$ then $X \cup_f D^n \simeq X \cup_g D^n$. So we have, with the previous notation, $$D^n \cup_{r} A \simeq D^n \cup_{s\circ h \circ r} A $$ Moreover $h$ induces a map $$\bar{h} \colon D^n \cup_{r} A \to D^n \cup_{h \circ r} B$$ ("extend" $h$ on $D^n$ to be the identity - so we are considering a map from the disjoint union- and then is easily verified that factors through the quotient), in the same fashion define $\bar{s}$. So the composition $\bar{h}\circ \bar{s}$ is a map $$ D^n \cup_{r} A \to D^n \cup_{h \circ r} B \to D^n \cup_{s \circ h \circ r} A \simeq D^n \cup_{r} A $$ But I cannot prove that the homotopy equivalence factors through the quotient, because it must induce another homotopy equivalence and I don't feel I have much control over the last homotopy equivalence. Some hints?