$AB$ and $XY$ are two chords of a circle whose centre is $P$. $A$ line $l$ passing through the centre $P$ bisects chords $AB$ and $XY$ both. Prove that $AB \parallel XY$.

(N/A) $1$. Let the circle have centre $P$. Let the line $l$ pass through $P$ and intersect $AB$ at point $M$ and $XY$ at point $N$.
$2$. Since the line $l$ bisects chord $AB$ at $M$,by the theorem 'the line drawn through the centre of a circle to bisect a chord is perpendicular to the chord',we have $PM \perp AB$. Thus,$\angle PMA = 90^{\circ}$.
$3$. Similarly,since the line $l$ bisects chord $XY$ at $N$,we have $PN \perp XY$. Thus,$\angle PNY = 90^{\circ}$.
$4$. Since $M, P,$ and $N$ lie on the same line $l$,the angles $\angle PMA$ and $\angle PNY$ are corresponding angles formed by the transversal $l$ intersecting lines $AB$ and $XY$.
$5$. Since $\angle PMA = 90^{\circ}$ and $\angle PNY = 90^{\circ}$,the corresponding angles are equal $(\angle PMA = \angle PNY = 90^{\circ})$.
$6$. Therefore,by the converse of the corresponding angles axiom,$AB \parallel XY$.