Prove that the line joining the mid-points of two parallel chords of a circle passes through the centre.

Explanation

Let `AB` and `CD` be two parallel chords having `P` and `Q` as their mid-points respectively. Let `O` be the centre of the circle.
Construction: Join `OP` and `OQ` and draw `OX∥AB` or `CD`
Now, `P` is the mid-point of `AB`
`OP⊥AB`, `∠BPO=90^@`
But `OX∥AB`
`therefore ∠XOQ=∠BPO` [Corresponding angles]
`∠XOQ=90^@`
Similarly, `Q` is the mid-point of `CD`
`OQ⊥CD`, `∠DQO=90^@`
But, OX∥CD
`therefore ∠POX=∠DQO=90^@` [Corresponding angles]
`∠POX+∠XOQ=90^@+90^@=180^@`
`POQ` is a straight line.
`therefore PQ` is a straight line passing through the centre of the circle.
Hence Proved.