The arcs intersected by these chords are not congruent.
Given that two circles with different radii have chords AB and CD, such that AB is congruent to CD.
Let r₁ and r₂ be the radii of two different circles with centers O and O' respectively.
Assuming that the each of the ∠АОВ and ∠CO'D is less than or equal to π.
Then, we have isosceles triangle AOB and CO'D such that,
AO = OB = r₁,
CO' = O'D = r₂,
Let us assume that r₁< r₂;
We can see that arc(AB) intersected by AB is greater than arc(CD), intersected by the chord CD;
arc(AB) > arc(CD) .......(1)
Indeed,
arc(AB) = r₁ angle (AOB)
arc(CD) = r₂ angle (CO'D)
So, we have to prove that ;
∠AOB >∠CO'D ......(2)
Since each angle is less than or equal to π, and so
∠AOB/2 and ∠CO'D/2 is less than or equal to π
it suffices to show that :
tan(AOB/2) >tan(CO'D/2) ......(3)
From triangle AOB :
tan(AOB/2) = AB/(2*r₁)
tan(CO'D/2) = CD/(2*r₂)
Since AB = CD and r₁ < r₂ (As obtained from the result of (3) ), therefore, arc(AB) > arc(CD).
Hence, for two circles with different radii have chords AB and CD, such that AB is congruent to CD but the arcs intersected by these chords are not congruent.
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