1. Clearly $B$, $C$, $B'$, $C'$ are concyclic, since they form an isosceles trapezoid. Applying the radical axis theorem to this circle and the circumcircles of $\triangle BB'P$, $\triangle CC'P$, we get that $A$, $P$, $P'$ are collinear. Similarly $A$, $Q$, $Q'$ are collinear. Since $AP \cdot AP' = AB \cdot AB' = AQ \cdot AQ'$, it follows that $P$, $Q$, $P'$, $Q'$ are concyclic.

2. Let $O_1$ be the other intersection point of the circumcircles of $\triangle BB'O'$ and $\triangle CC'O'$. As in 1., $A$, $O'$, $O_1$ are collinear, and $Q$, $Q'$, $O'$, $O_1$ are concyclic.

3. Observe that $\triangle ABC$ and $\triangle AB'C'$ are symmetric with respect to the internal angle bisector of $\angle A$, and $\angle B'C'A = \angle B$, so we get
$\angle OAC = 90^\circ - \angle B = 90^\circ - \angle B'C'A$, i.e. $AO \perp B'C'$. Using the symmetry again, we know that
$AO' \perp BC$. Therefore $AO'O_1 \parallel PQ$.

4. Let $a$, $b$, $c$, $R$, $\triangle h_a$ be the three side lengths, the circumradius, the area of the triangle, and the $A$-altitude respectively. Since $AO' \cdot AO_1 = AB \cdot AC$,
$$AO_1 = \frac{AB \cdot AC}{AO'} = \frac{bc}{R} = \frac{bc \sin A}{R \sin A} = \frac{4\Delta}{a} = 2h_a.$$
Hence the reflection of $A$ with respect to line $BC$ is $O_1$. Therefore $PQAO_1$ is an isosceles trapezoid.

5. By 1. and 3., we know $\angle QQ'P' = \angle QPP' = \angle PAO_1$. Also by 2. and 4., we know $\angle QQ'O' = \angle QO_1A = \angle PAO_1$. From this it is proved that $O'$, $P'$, $Q'$ are collinear.
Since $\angle AQ'P' = \angle QQ'P' = \angle PAO_1 = \angle P'AO_1$, we get that $O'A$ is tangent to the circumcircle of $\triangle AP'Q'$ at $A$, so $O'P' \cdot O'Q' = O'A^2 = OA^2$. Q.E.D.