Question

Show that in a circle the greater of two chord is nearer tothe centre

Answer 1

Given: AB and CD are two chords of a circle C(O, r). OP < OQ, where OP ⊥ CD and OQ ⊥ AB
To prove: CD > AB
Proof:
AQ = ½ AB            (Perpendicular from the centre of a circle to a chord bisects the cord)
CP = ½ CD                (Perpendicular from the centre of a circle to a chord bisects the cord)
In right ΔAOQ,
⇒ AO2 = OQ2 + AQ2
⇒ AQ2 = AO2 - OQ2 ....(1)

In right ΔCOP,
⇒ CO2 = CP2 + OP2
⇒ CP2 = CO2 - OP2 ....(2)

⇒ OP < OQ
⇒ OP2 < OQ2
⇒ -OP2 < -OQ2
⇒ CO2 - OP2 < AO2 - OQ2 (CO = AO)
⇒ CP2 > AQ2        (from (1) and (2))
⇒ ¼ CD2 > ¼ AB2
⇒ CD2 > AB2
⇒ CD > AB