In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.
is a diameter of a circle with centre O. AB and CD are two chords such that AB || CD.
AB = CD
Draw OL ⊥ AB and OM ⊥ CD.
In ΔOLB and ΔOMC, we have:
∠OLB = ∠OMC [90° each]
∠OBL = ∠OCD [Alternate angles as AB || CD]
OB = OC [Radii of a circle]
∴ ΔOLB ≅ ΔOMC (AAS criterion)
Thus, OL = OM (CPCT)
We know that chords equidistant from the centre are equal.
Hence, AB = CD