Question

Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other.

Answer 1

Given:

1. O is the centre of the circle.
2. seg PQ is the diameter.
3. Diameter PQ bisects the chords AB and CD at the points M and N respectively.

To prove: chord AB || chord CD.

Proof:

Point M is the midpoint of chord AB.      ...(Given)

∴ seg OM ⊥ chord AB      ...(The line segment joining the center of the circle and the midpoint of the chord is perpendicular to the chord.)

∴ ∠OMA = 90°      ...(i)

Point N is the midpoint of chord CD.      ...(Given)

∴ seg ON ⊥ chord CD      ...(The line segment joining the center of the circle and the midpoint of the chord is perpendicular to the chord.)

∴ ∠ONC = 90°      ...(ii)

Now, ∠OMA + ∠ONC

= 90° + 90°      ...[From (i) and (ii)]

∴ ∠OMA + ∠ONC = 180°

But, ∠OMA and ∠ONC form a pair of interior angles on lines AB and CD when ∠MN is their transversal.

∴ chord AB || chord CD      ...(Interior angles test)