Question

two circles with Centre A and B intersect at X and Y. PQ is a chord of one circle parallel to XY. prove that AB or AB produced bisects PQ

Answer 1

(Notice that I'm a little bad at drawing but I hope that is comprehensible)
Without loss of generality (by symmetry):
In ΔAXB and ΔAYB,
AX = AY (Radii of the same circle)
BX = BY (Radii of the same circle)
AB = AB (Common)
∴ ΔAXB ≅ ΔAYB (SSS criterion)
⇒ ∠4 = ∠5 (CPCT) ---(1)
In AOX and ΔAOY,
∠4 = ∠5 (From 1)
AX = AY (Radii of the same circle)
AO = AO (Common)
∴ ΔAOX ≅ ΔAOY (SAS criterion)
⇒∠1 = ∠2 (CPCT) 
But ∠1 + ∠2 = 180° (Linear Pair Axiom)
2∠1 = 180°
⇒ ∠1 = 90°
⇒ ∠2 = 90°
Now, CD║XY
Therefore ∠1 = ∠3 (Alternate interior angles)
Or ∠3 = 90° 
Since AM⊥PQ and AM passes through B (The center), 
⇒PM=PQ (A perpendicular from the center to a chord bisects the chord)
Hence Proved

Answer 2

....... it will help you