In the adjoining figure, OD is perpendicular to the chord AB of a circle whose centre is O. If BC is a diameter, show that AC = 2 OD and AC ||OD.
   

AB is the chord and OD ⊥ AB ⇒ BD = AD
In ΔABC, D is the mid-point of AB and O is the mid-point of BC
∴ OD || AC
Now in, ΔABC, OD || AC
\\         I By BPT Theorem
or  AC = 2DO = 2OD
∴ AC = 2 OD and AC || OD.