= Solution
If every <monomorphism> is strong, let $m:A\to B$ be both a <monomorphism> and an <epimorphism>. Use $m$ as both vertical maps in the lifting square, with horizontal maps $1_A$ and $1_B$. The <strong monomorphism> property gives $d:B\to A$ satisfying $dm=1_A$ and $md=1_B$. Thus $m$ is an <isomorphism>, proving that the <category> is <balanced>.
For the converse, take $e:X\to Y$ epic, $m:A\to B$ monic, and $mu=ve$. Form the <pullback in a category> of $m$ along $v$:
$$
P=Y\times_B A,\qquad p:P\to Y,\quad q:P\to A,\quad mq=vp.
$$
The equality $mu=ve$ gives $t:X\to P$ with $pt=e$ and $qt=u$. By preservation of <monomorphisms> under <pullback in a category>, $p$ is monic. It is also epic: if $hp=kp$, then $he=hpt=kpt=ke$, and cancellation of $e$ gives $h=k$. The <balanced category> hypothesis therefore makes $p$ an <isomorphism>.
Set $d=qp^{-1}$. Then $md=v$ and $de=qt=u$. Any other such lift agrees with $d$ after composing with the <monomorphism> $m$, so it is equal to $d$. Consequently \b[in a <category> with <pullbacks in a category>, being <balanced> is equivalent to every <monomorphism> being a <strong monomorphism>]. This is the <balanced categories with pullbacks have strong monomorphisms> principle; the proof does not assume pullback stability of epimorphisms.
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