Question

Hey folks! ;D

Explain this theorem from the chapter circles : Chords equidistant from the centre of the circle are equal in length.

Draw a diagram, for better understanding.

Thanks in advance! :D

Answer 1

$\bf{\huge{\underline{\boxed{\tt{\blue{Solution}}}}}}$

In the above diagram we have: 

• A circle with center O 

• PQ and RS are two chords of circle 

• OA and OB is the distance of chords PQ

• RS respectively from the center of circle 

• OA is equal to OB 

Now,

As per the property 6 of circle

"Chords equidistant from the center of circle are equal in length", we get: 

PQ = RS

Answer 2

$\huge\underline\mathfrak\red{Statement-}$

Chords equidistant from the centre of the circle are equal in length.

$\huge\underline\mathfrak\red{Explanation-}$

Given :

• AB and CD are two chords of a circle having center O.
• OL = OM

To prove :

• AB = CD

Construction :

• Join OA and OC

Proof :

In ∆OLA and ∆OMC

$\implies$ OL = OM ( given )

$\implies$ ∠OLA = ∠OMC ( Each 90° )

$\implies$ OA = OC ( Radii of same circle are equal )

$\therefore$ ∆OLA ≅ ∆OMC

$\implies$ LA = MC ( Corresponding parts of congruent ∆s are equal ) __________(1)

Also, we know that a perpendicular drawn from the center of the circle bisects the chord.

$\therefore$ LA = LB and MC = MD

______________

From (1),

LA = MC

Twice both sides,

2LA = 2MC

$\implies$ LA + LA = MC + MC ________(2)

Also, LA = LB and MC = MD _______(3)

Put the value of (3) in (2),

$\implies$ LA + LB = MC + MD

( LA + LB = AB and MC + MD = CD )

$\orange{\boxed{\pink{\underline{\underline{\red{\mathcal{AB=CD}}}}}}}$

Hence proved!