Question

Prove that, of any two chords of a circle, the greater chord is nearer to the 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 = $\frac{1}{2}$ AB (Perpendicular from the centre of a circle to a chord bisects the cord)
CP = $\frac{1}{2}$ 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))
⇒$\frac{1}{4}$ CD² > $\frac{1}{4}$ AB²
⇒ CD² > AB²
⇒ CD > AB