Two circles with centres $O$ and $O'$ intersect at two points $A$ and $B$. $A$ line $PQ$ is drawn parallel to $OO'$ through $A$ (or $B$) intersecting the circles at $P$ and $Q$. Prove that $PQ = 2 OO'$.

(N/A) Given: Two circles with centres $O$ and $O'$ intersect at $A$ and $B$. $A$ line $PQ$ is drawn parallel to $OO'$ passing through $A$,intersecting the circles at $P$ and $Q$.
To prove: $PQ = 2 OO'$.
Construction: Draw $OC \perp PA$ and $O'D \perp AQ$.
Proof:
$1$. Since the perpendicular from the centre of a circle to a chord bisects the chord,we have:
$PA = 2 CA$ (as $OC \perp PA$) $...(1)$
$AQ = 2 AD$ (as $O'D \perp AQ$) $...(2)$
$2$. Adding equations $(1)$ and $(2)$:
$PA + AQ = 2 CA + 2 AD$
$PQ = 2(CA + AD)$
$3$. Since $PQ \parallel OO'$,$OC \perp PQ$,and $O'D \perp PQ$,the quadrilateral $CDO'O$ is a rectangle.
Therefore,$CD = OO'$.
$4$. Substituting $CD = OO'$ in the equation $PQ = 2(CA + AD)$:
$PQ = 2 CD = 2 OO'$.
Hence,$PQ = 2 OO'$ is proved.