Theorem 10.6 Class 9 Maths Explanation with Proof

Today, we will discuss theorem 10.6 class 9 Maths which is related to Chapter 10 Circles Class 9 Mathematics. After understanding theorem 10.6, you can solve the exercise questions given in the NCERT book of Class 9 Maths.

 Theorem 10.6 Class 9

Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres)

Given 

A circle with centre O and two equal chords AB and PQ., i.e. AB=PQ.

To Prove

AB and PQ are equidistant from the centre O.

Construction 

Draw OM⊥AB and ON⊥PQ. Join OA, OB, OP and OQ.

Proof 

We know that a perpendicular from the centre to a chord bisects the chord.

`AM=MB=frac{1}{2} AB`          (∵OM⊥AB)

And  `PN=NQ= frac{1}{2} PQ`   (∵ON⊥PQ)

∵    chord AB=chordPQ                (given)

⇒  `frac{1}{2} AB=frac{1}{2} PQ`

 

Then AM=MB=PN=NQ

Now, in ∆OMA and ∆ONQ 

OA=OQ                            (Radii of the same circle)

AM=NQ                           (proved above)

∠OMA=∠ONQ                (each 90°)

∴∆OMA=∆ONQ         (By RHS congruence rule)

Then, OM=ON           (By CPCT)

Hence, PQ and AB are equidistant from the centre O.  

Hence proved

 Related Topics

1. Theorem 10.7

2. Theorem 10.5