$AB$ and $CD$ are two parallel chords of a circle with centre $P$. $l$ is the perpendicular bisector of chord $AB$. Prove that $l$ bisects chord $CD$.

(N/A) $1$. Let the circle have center $P$. Since $l$ is the perpendicular bisector of chord $AB$,it must pass through the center $P$ of the circle.
$2$. We are given that $AB \parallel CD$.
$3$. Since $l$ is perpendicular to $AB$ $(l \perp AB)$ and $AB \parallel CD$,it follows that $l$ must also be perpendicular to $CD$ $(l \perp CD)$.
$4$. $A$ line passing through the center of a circle and perpendicular to a chord bisects the chord.
$5$. Since $l$ passes through the center $P$ and is perpendicular to $CD$,$l$ must bisect chord $CD$.