Question

Prove that the lengths of all chords at equal distances from the centre of a circle are equal.

Answer 1

Given: Chords AB and CD which are at an equal distance from the centre, O of the circle

To prove: AB = DC

Construction: Join AO and OD.

Proof:

OF and OE are the respective perpendicular distances of the chords AB and DC from the centre of the circle.

OF = OE (Given)

In ΔOAF and ΔODE, OF ⊥ AB and OE ⊥ DC

By Pythagoras theorem, we have:

OA² = OF² + FA²

⇒ FA² = OA² − OF² ... (1)

Also, OD² = DE²+ OE²

⇒ DE² = OD² − OE² ... (2)

However, OA = OD = Radii of the same circle

Also, OF = OE

So, from equations (1) and (2):

FA²= DE²

⇒ FA = DE

We know that perpendicular from the centre of the circle to the chord bisects the chord.

⇒ FA = FB and DE = EC

Now, AB = AF + FB

= 2AF

= 2DE

= DC (DE = )

Thus, the lengths of all chords at equal distances from the centre of the circle are equal.