$Q$ in $A$ is a quaternion $F$-algebra $Q'$, and we have $$A\simeq Q\otimes_FQ'.$$ This relation holds if and only if $A\otimes_FQ$ is Brauer-equivalent to $Q'$, hence (3) follows.

Let $\sigma$ be an orthogonal involution on $A$. Then $(A,\sigma)_{F_Q}$ is hyperbolic if and only if $(A,\sigma)$ is hyperbolic or contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$.

As noticed in the introduction, any $(A,\sigma)$ containing ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$ is hyperbolic over $F_Q$, so we only have to prove the converse. Assume $(A,\sigma)_{F_Q}$ hyperbolic. Since $F$ is quadratically closed in $F_Q$, the discriminant of $\sigma$ is trivial, so $C(A,\sigma)$ is a direct product of two quaternion algebras. At least one of these quaternion algebras splits over $F_Q$ by Proposition \[deg4orth.prop\](1), hence that component must be either split or isomorphic to $Q$. It follows that $(A,\sigma)$ is either hyperbolic or contains ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, by Proposition \[deg4orth.prop\].

Note that a central simple algebra of degree $4$ with hyperbolic involution does not necessarily contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, even if it contains $Q$: if $Q'$ is a quaternion $F$-algebra such that $Q\otimes_FQ'$ has index $2$, then $(A,\sigma)=(M_2(F),{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})\otimes(Q',{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})$ is hyperbolic over $F$ and satisfies $C(A,\sigma)\simeq M_2(F)\times Q'$ by [@KMRT (15.12)]. Therefore, Proposition \[deg4orth.prop\](2) shows that $(A,\sigma)$ does not contain ${(Q,{\overline{\rule{2.5mm}{0mm}\rule{0mm}{4pt}}})}$, even though Proposition \[deg4orth.prop\](3) shows that $A$ contains 