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Home >> Circle >> Properties of Circle >> Chords equidistant from the center of circle are equal in length >>

Chords equidistant from the center of circle are equal in length

Equal chords subtend equal angles at the center of a circle If Angles subtended by the chords at the center of circle are equal, then chords are also equal Perpendicular from the center of a circle to a chord bisects the chord Line drawn from the center of circle to bisect a chord, is perpendicular to the chord Equal chords are equidistant from the center of circle
Chords equidistant from the center of circle are equal in length

Before you understand the property "Chords equidistant from the center of circle are equal in length", you are advised to read:

What is Center of Circle ?
What are the Chords of Circle ?
What is Distance of a Line from the point ?

Observe the following diagram:



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 i.e."Chords equidistant from the center of circle are equal in length", we get:
PQ = RS

How to prove this property : Chords equidistant from the center of circle are equal in length

Before you prove this property, you are advised to read:

What is RHS Rule of congruency in Triangles ?
What are Corresponding Parts of Congruent Triangles ?
What is Radii of Circle ?
What is Distance of a Line from the point ?
Property : Perpendicular from the center of a circle to a chord bisects the chord

Observe the following diagram:



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, observe carefully:
Since OA and OB is the distance of chords PQ & RS respectively from the center of circle.
And as per the property which says "The length of perpendicular from a point to a line is the distance of line from the point", so we get:
OA is perpendicular to PQ and OB is perpendicular to RS (as shown below) ..... (statement 1)



Also, we know that:
O is the center of circle (given)
And OA is perpendicular to PQ (proved in above statement 1)
So apply Property 3 of circle, "The perpendicular from the center of a circle to a chord bisects the chord" and we get:
AQ = PA
Or we can write it as:
2AQ = PQ ..... (statement 2)

Similarly, O is the center of circle (given)
And OB is perpendicular to RS (proved in above statement 1)
So apply Property 3 of circle, "The perpendicular from the center of a circle to a chord bisects the chord" and we get:
RB = BS
Or we can write it as:
2RB = RS ..... (statement 3)

Now, join points O & Q and O & R (as shown below):
This would give us two triangles i.e.
△ AOQ and △ BOR (as highlighted below):



△ AOQ ^ △ BOR

Angle 1 = Angle 2 (90 degree each - proved in above statement 1)
OQ = OR (radii of circle are always equal)
OA = OB (given)
Therefore, on applying RHS Rules of congruency, we get:
△ AOQ ^ △ BOR

Since, we know that corresponding parts of congruent triangles are equal, so we get:
AQ = RB

Multiply both sides by 2 and we get:
2AQ = 2RB

Put the values from above statement 2 & 3 and we get:
PQ = RS

Hence, this proves property 6 of circle i.e. "Chords equidistant from the center of circle are equal in length"

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